Openstax College Physics Instructor Solutions Manual Chapter 13 CHAPTER 13: TEMPERATURE, KINETIC THEORY, AND THE GAS LAWS 13.1 TEMPERATURE 1. What is the Fahrenheit temperature of a person with a 39.0C fever? Solution 9 T F (TC ) 32.0 5 9 T F (39.0C) 32.0C 102F 5 2. Frost damage to most plants occurs at temperatures of 28.0F or lower. What is this temperature on the Kelvin scale? Solution To convert from Fahrenheit to Kelvin, first convert to Celsius: TC 5 5 (T F 32.0) (28.0F 32.0F) 2.22C 2.2C 9 9 Then convert to Kelvin: TK 2.2 273.15 271.0 K 3. Solution To conserve energy, room temperatures are kept at 68.0F in the winter and 78.0F in the summer. What are these temperatures on the Celsius scale? TSummer 4. Solution 5 (T F 32.0F) 9 5 (T F 32.0F) 9 TWinter 5 (68.0F 32.0F) 20.0C 9 5 (78.0F 32.0F) 25.6C 9 A tungsten light bulb filament may operate at 2900 K. What is its Fahrenheit temperature? What is this on the Celsius scale? This is a two step process. First we convert Kelvin to Celsius then to Fahrenheit: Openstax College Physics Instructor Solutions Manual Chapter 13 TC TK 273.15 2900K 273.15 2626.85C = 2600C 9 9 T F TC 32.0 (2626.85C) 32.0C 4760F = 4800 F 5 5 5. The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale? Solution This is a two step process but we can do it in one step by combining the conversion equations for Kelvin to Celsius and Celsius to Fahrenheit. 9 9 T F (TK 273.15) 32.0 (5750 K 273.15) 32.0F 9890F 5 5 6. One of the hottest temperatures ever recorded on the surface of Earth was 134F in Death Valley, CA. What is this temperature in Celsius degrees? What is this temperature in Kelvin? Solution TC 7. (a) Suppose a cold front blows into your locale and drops the temperature by 40.0 Fahrenheit degrees. How many degrees Celsius does the temperature decrease when there is a 40.0F decrease in temperature? (b) Show that any change in temperature in Fahrenheit degrees is nine-fifths the change in Celsius degrees. Solution 5 5 (a) ΔTC TC2 TC1 (T F 2 32) (T F1 32) 9 9 5 5 5 (T F 2 T F1 ) T F (40) 22.2 C 9 9 9 5 5 (T F 32.0) (134F 32.0) 56.7C 57C 9 9 TK TC 273.15 57C 273.15 330 K 9 (b) We know that TF TF2 TF1 . We also know that T F2 TC2 32 and 5 9 9 9 T F1 TC1 32 . So, substituting, we have ΔT F TC2 32 TC1 32 . 5 5 5 9 Partially solving and rearranging the equation, we have T F TC2 TC1 . 5 9 Therefore, T F TC . 5 8. (a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value? Openstax College Physics Solution Instructor Solutions Manual Chapter 13 9 (a) To answer this we need to set T F TC T , using T F TC 32 so 5 4 9 T T 32.0. Solving gives: T 32.0 , or T 40.0 . At this temperature, 5 5 the Fahrenheit and Celsius scales have the same value. (b) For this part, we need to let TK TC 273.15 TF T . Using an equation from 9 part (a), T F TC 32 , and substituting in TC TK 273.15 along with 5 TK T TF , we have: 9 4 (T 273.15) 32.0 T 459.67 5 5 T 575F or 575K T 13.2 THERMAL EXPANSION OF SOLIDS AND LIQUIDS 9. The height of the Washington Monument is measured to be 170 m on a day when the temperature is 35.0C . What will its height be on a day when the temperature falls to – 10.0C ? Although the monument is made of limestone, assume that its thermal coefficient of expansion is the same as marble’s. Solution Using Table 13.2 to find the thermal coefficient of expansion of marble: L L0 L L0 1 T 170 m 1 7 106 / C 45.0C 169.95 m (Answer rounded to five significant figures to show the slight difference in height.). 10. How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by 15C ? Its original height is 321 m and you can assume it is made of steel. Solution Using Table 13.2 to find the thermal coefficient of expansion of steel: L LT (1.2 10 5 /C)(321 m)(15 C) 0.058 m 11. What is the change in length of a 3.00-cm-long column of mercury if its temperature changes from 37.0C to 40.0C , assuming the mercury is unconstrained? Solution Using Table 13.2 to find the thermal coefficient of expansion of mercury: L LT (6.0 10 5 / C)(0.0300 m)(3.00C) 5.4 10 6 m 12. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 35.0C greater than when they were laid? Their original length is 10.0 m. Openstax College Physics Instructor Solutions Manual Chapter 13 Solution Using Table 13.2 to find the thermal coefficient of expansion of steel: L LT (1.2 10 5 /C)(10.0 m)(35.0 C) 0.0042 m 13. You are looking to purchase a small piece of land in Hong Kong. The price is “only” $60,000 per square meter! The land title says the dimensions are 20 m 30 m . By how much would the total price change if you measured the parcel with a steel tape measure on a day when the temperature was 20C above normal? Solution On the warmer day, our tape measure will expand linearly. Therefore, we will measure each dimension to be smaller than its actual size. (Note that the parcel of land does not actually change, so we do not use the equation for thermal expansion in two dimensions.) Calling these new dimensions l and w , we will find a new area, A . Let’s calculate these new dimensions: 1.2 10 5 l l 0 l (20 m) (20C)(20 m)( ) = 19.9952 m C 1.2 10 5 w = w0 w = (30 m) (20C)(30 m)( ) = 29.9928 m C A = l w = (29.9928 m)(19.9952 m) = 599.71 m 2 $60,000 2 $60,000 ) (( 600 599 . 71 ) m )( ) $17,000 m2 m2 Because the area gets smaller, the price of the land DECREASES by ~$17,000. Cost Change = ( A A)( 14. Global warming will produce rising sea levels partly due to melting ice caps but also due to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water 1.00 km high for a temperature increase of 1.00C. Note that this calculation is only approximate because ocean warming is not uniform with depth. Solution To deal with volume increase, we need to use the equation L L0 T . (We need 3 to divide by three because we are looking for the change in length, not volume.) Use Table 13.2 to find water’s coefficient of volume expansion: L 3 L0 T (7.0 10 5 / C)(1 103 m)(1.00C) 7 10 3 m 15. Show that 60.0 L of gasoline originally at 15.0C will expand to 61.1 L when it warms to 35.0C, as claimed in Example 13.4. Solution V ' V0 VT 60.0 L (9.50 10 4 / C)(60.0 L)(20.0 C) 61.1 L Openstax College Physics 16. Instructor Solutions Manual Chapter 13 (a) Suppose a meter stick made of steel and one made of invar (an alloy of iron and nickel) are the same length at 0C . What is their difference in length at 22.0C ? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes. Solution (a) L ( s i ) LT (1.2 10 5 / C 9.0 10 7 / C)(1.0 m)(22 C) 2.4 10 4 m (b) L ( s i ) LT (1.2 10 5 / C 9.0 10 7 / C)(30.0 m)(22 C) 7.3 10 3 m 17. (a) If a 500-mL glass beaker is filled to the brim with ethyl alcohol at a temperature of 5.00C, how much will overflow when its temperature reaches 22.0C ? (b) How much less water would overflow under the same conditions? Solution Assume that the glass beaker is initially at room temperature ( 22.0C ), so that its temperature does not change. (a) Ethyl alcohol: V VT (1.100 10 3 / C)(0.500 L)(17.0 C) 9.35 10-3 L 9.35 mL (b) Water: V VT (2.10 10 4 /C)(0.500 L)(17.0 C) 1.79 10 3 L 1.79 mL 18. Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at 10.0C. What volume of radiator fluid will overflow when the radiator and fluid reach their 95.0C operating temperature, given that the fluid’s volume coefficient of expansion is 400 10 –6 /C ? Note that this coefficient is approximate, because most car radiators have operating temperatures of greater than 95.0C. Solution V ( rf c )VT (4.00 10 4 / C 5.110 5 / C)(16.0 L)(85.0 C) 0.475 L 19. A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show that this decrease cannot be due to thermal 3 contraction by calculating the decrease in level if the 350 cm of coffee is in a 7.00cm-diameter cup and decreases in temperature from 95.0C to 45.0C. (Most of the drop in level is actually due to escaping bubbles of air.) Openstax College Physics Instructor Solutions Manual Chapter 13 V πr 2 h Solution V πr 2 h ( w g )VT πr 2 h h ( w g )VT πr 2 2.10 10 4 / C 2.7 10 5 / C 3.50 10 4 m 3 50.0C 3.50 10 m 2 2 4 8.32 10 m 0.832 mm The work shows that the change in volume (from thermal expansion) of the coffee and the cup does not account for all of the level drop. Thus, other factors must be held accountable (such as escaping bubbles of air). 20. (a) The density of water at 0C is very nearly 1000 kg/m 3 (it is actually 999.84 kg/m 3 ), whereas the density of ice at 0C is 917 kg/m 3 . Calculate the pressure necessary to keep ice from expanding when it freezes, neglecting the effect such a large pressure would have on the freezing temperature. (This problem gives you only an indication of how large the forces associated with freezing water might be.) (b) What are the implications of this result for biological cells that are frozen? Solution (a) If we start with the freezing of water, then it would expand to 1000 kg/m 3 (1 m 3 )( ) = 1.09 m 3 of ice. 3 917 kg/m 0.0900m 3 V B( )( 2.20 10 9 N/m 2 ) 1.98 10 8 N/m 2 3 V0 1.00m (b) Biological cells that are frozen must be left free to expand when they are unfrozen, otherwise they will undergo extreme pressures and probably be destroyed. In the freezing process, the cells experience large outward pressures. If they don’t have room to expand, they will likely be destroyed. P 21. Show that 3 , by calculating the change in volume V of a cube with sides of length L. Solution From the equation DL = aL0 DT we know that length changes with temperature. We also know that the volume of a cube is related to its length by V = L . Using the equation V = V0 + DV and substituting for the sides we get V = ( L0 + DL) 3 . Then we 3 replace DL with DL = aL0 DT to get V ( L0 L0 T ) 3 L30 (1 T ) 3 . Since a D T is small, we can use the binomial expansion to get V L30 (1 3T ) L30 3L30 T . Rewriting the length terms in terms of volume gives V = V0 + DV = V0 + 3aV0 DT . By comparing forms we get DV = bV0 DT = 3aV0 DT . Thus, 3 . Openstax College Physics Instructor Solutions Manual Chapter 13 13.3 THE IDEAL GAS LAW 22. Solution 5 2 The gauge pressure in your car tires is 2.50 10 N/m at a temperature of 35.0C when you drive it onto a ferry boat to Alaska. What is their gauge pressure later, when their temperature has dropped to –40.0C ? P1 2.50 10 5 N/m 2 1.013 10 5 N/m 2 3.513 10 5 N/m 2 P2V2 NkT2 P1V1 NkT1 since V1 = V2 PT 233.15 K P2 1 2 (3.153 10 5 N/m 2 )( ) 2.658 10 5 N/m 2 T1 308.15 K P 2.658 10 5 N/m 2 1.013 10 5 N/m 2 1.64 10 5 N/m 2 1.62 atm 23. Solution 5 2 2 Convert an absolute pressure of 7.00 10 N/m to gauge pressure in lb/in . (This 2 value was stated to be just less than 90.0 lb/in in Example 13.9. Is it?) Pg Pa 1 atm 7.00 10 5 N/m 2 1.013 10 5 N/m 2 5.987 10 5 N/m 2 Pg (5.987 10 5 N/m 2 ) ( 24. 1.00 lb/in. 2 ) 86.8 lb/in. 2 6.895 10 3 N/m 2 Suppose a gas-filled incandescent light bulb is manufactured so that the gas inside the bulb is at atmospheric pressure when the bulb has a temperature of 20.0C . (a) Find the gauge pressure inside such a bulb when it is hot, assuming its average temperature is 60.0C (an approximation) and neglecting any change in volume due to thermal expansion or gas leaks. (b) The actual final pressure for the light bulb will be less than calculated in part (a) because the glass bulb will expand. What will the actual final pressure be, taking this into account? Is this a negligible difference? Solution (a) P1V NkT1 , P2V NkT2 P2 P1T2 333.15K 5 2 5 (1.013 10 5 N/m 2 ) 1.15110 N/m 1.15110 Pa T1 293.15K Pg P2 1 atm 1.15110 5 N/m 2 1.013 10 5 N/m 2 1.38 10 4 N/m 2 0.136 atm (b) First, let P2 = the final pressure without glass bulb expansion, P2' = the final pressure with glass bulb expansion, and T2 = the final temperature of the light bulb. Also, let V ' = the final volume pressure with glass bulb expansion. Openstax College Physics Instructor Solutions Manual Chapter 13 P2V NkT2 P (V V ) NkT2 ' 2 P2 P2 V ) V V (V V ) / V 1 (V / V ) V VT P2' P2 ( V T V PT 333.15K P2 1 2 (1.013 10 5 N/m 2 )( ) 1.151 10 5 Pa T1 293.15K P2' P2 1.151 10 5 N/m 2 1.150 10 5 N/m 2 1 T 1 (2.7 10 5 / C)(40.0C) Pg' P2' 1.00 atm 1.150 10 5 N/m 2 1.013 10 5 N/m 2 1.37 10 4 N/m 2 0.135 atm The difference between this value and the value from part (a) is negligible. 25. Large helium-filled balloons are used to lift scientific equipment to high altitudes. (a) What is the pressure inside such a balloon if it starts out at sea level with a temperature of 10.0C and rises to an altitude where its volume is twenty times the original volume and its temperature is –50.0C ? (b) What is the gauge pressure? (Assume atmospheric pressure is constant.) Solution (a) P2V2 NkT2 P1V1 NkT1 P2 P1V1T2 P1V1T2 (1.013 10 5 N/m 2 )( 223.15 K ) V2T1 20V1T1 20(283.15 K) 3.92 10 3 N/m 2 3.94 10 2 atm (b) Gauge pressure = 3.94 102 atm 1.00 atm 0.961 atm 26. Confirm that the units of nRT are those of energy for each value of R : (a) 8.31 J/mol K , (b) 1.99 cal/mol K , and (c) 0.0821 L atm/mol K . Solution (a) nRT (mol)( J )(K) J mol K cal )(K) cal (b) nRT (mol)( mol K L atm )(K) L atm (m 3 )(N/m 2 ) N m J (c) nRT (mol)( mol K Openstax College Physics Instructor Solutions Manual Chapter 13 27. In the text, it was shown that N / V 2.68 1025 m3 for gas at STP. (a) Show that this quantity is equivalent to N / V 2.68 1019 cm3 , as stated. (b) About how many atoms are there in one μm3 (a cubic micrometer) at STP? (c) What does your answer to part (b) imply about the separation of atoms and molecules? Solution N 2.68 10 25 1 m 19 3 (a) 2.68 10 cm 3 V m 100 cm 3 3 N 2.68 10 25 1m 2.68 10 7 m -3 (b) 3 6 V m 1.00 10 m (c) This says that the average volume of atoms and molecules must be on the order of N 1 3.73 10 8 m 3 7 -3 2.68 10 m 2.68 10 7 m -3 Or the average length of an atom is less than approximately (3.73 108 m3 )1/ 3 3.34 103 m 3 nm. Since atoms are widely spaced, the average length is probably more on the order of 0.3 nm. V 28. Solution 29. Solution Calculate the number of moles in the 2.00-L volume of air in the lungs of the average person. Note that the air is at 37.0C (body temperature). PV nRT n PV (1.013 105 N/m 2 )(2.00 10 3 m 3 ) 7.86 10 2 mol RT (8.31 J/mol K)(310.15 K) 3 An airplane passenger has 100 cm of air in his stomach just before the plane takes off from a sea-level airport. What volume will the air have at cruising altitude if cabin 4 2 pressure drops to 7.50 10 N/m ? P1V1 P2V2 V2 P1 1.013 105 N/m 2 V1 (100 cm 3 ) 135 cm 3 4 2 P2 7.50 10 N/m 30. (a) What is the volume (in km 3 ) of Avogadro’s number of sand grains if each grain is a cube and has sides that are 1.0 mm long? (b) How many kilometers of beaches in length would this cover if the beach averages 100 m in width and 10.0 m in depth? Neglect air spaces between grains. Solution 1 mm 3 1 km 5 3 14 3 6 (a) 6.02 10 grains 6.02 10 km 6.02 10 m grain 10 mm V (6.02 1014 m3 ) (b) LWD V L 6.02 1011 m 6.02 108 km WD (100 m)(10.0 m) 3 23 Openstax College Physics 31. Instructor Solutions Manual Chapter 13 –7 2 An expensive vacuum system can achieve a pressure as low as 1.00 10 N/m at 20C . How many atoms are there in a cubic centimeter at this pressure and temperature? Solution PV (1.00 10 7 N/m 2 )(1.00 10 6 m 3 ) PV NkT N 2.47 10 7 atoms 23 kT (1.38 10 J/K)(293.1 5 K) 32. The number density of gas atoms at a certain location in the space above our planet is about 1.00 1011 m3 , and the pressure is 2.75 10 –10 N/m 2 in this space. What is the temperature there? Solution P 1 (2.75 10 10 N/m 2 ) 199 K 73.9C 11 3 23 N ( 1 . 00 10 / m ) ( 1 . 38 10 J/k) ( )k V (which seems reasonable) 33. 5 2 A bicycle tire has a pressure of 7.00 10 N/m at a temperature of 18.0C and contains 2.00 L of gas. What will its pressure be if you let out an amount of air that 3 has a volume of 100 cm at atmospheric pressure? Assume tire temperature and volume remain constant. Solution P1V (7.00 105 N/m 2 )( 2.00 10 3 m3 ) 3.484 10 23 23 kT (1.38 10 J/K)(291.1 5 K) Then, we need to determine how many molecules were removed from the tire: 10 6 m 3 (1.013 105 N/m 2 )100 cm 3 cm 3 PV PV NkT N 2.52110 21 23 kT (1.38 10 J/K)(291.1 5K) PV NkT T P1V N1kT N1 We can now determine how many molecules remain after the gas is released: N 2 N1 N 3.484 1023 2.5211021 3.459 1023 Finally, the final pressure is: N 2 kT (3.459 10 23 )(1.38 10 23 J/K )( 291.15 K ) V 2.00 10 3 m 3 6.95 10 5 N/m 2 6.95 10 5 Pa P2 Openstax College Physics 34. Instructor Solutions Manual Chapter 13 A high-pressure gas cylinder contains 50.0 L of toxic gas at a pressure of 1.40 107 N/m 2 and a temperature of 25.0C . Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature (–78.5C) to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? (b) What is the final pressure if one-tenth of the gas escapes? (c) To what temperature must the tank be cooled to reduce the pressure to 1.00 atm (assuming the gas does not change phase and that there is no leakage during cooling)? (d) Does cooling the tank appear to be a practical solution? Solution (a) Assume N 2 » N1 , then since P1V = N1 KT1 and P2V = N 2 KT2 P2 P1 ( 194.65K T2 0.914 10 7 N/m 2 9.14 10 6 Pa ) (1.40 10 7 N/m 2 ) T1 298.15K (b) So, we now we know that N 2 0.9 N1 , and T N 194.65K P2 P1 2 2 (1.40 10 7 N/m 2 )(0.900) 298.15K T1 N 1 0.823 10 7 N/m 2 8.23 10 6 Pa P 1.013 105 N/m 2 P2 T2 2.16 K T2 T1 2 (298.15 K) 7 2 P1 T1 1.40 10 N/m P1 (d) No. The final temperature needed is much too low to be easily achieved for a large object. (c) 35. Solution 36. 7 2 Find the number of moles in 2.00 L of gas at 35.0C and under 7.41 10 N/m of pressure. PV nRT n PV (7.4110 7 N/m 2 )( 2.00 10 3 m 3 ) 57.9mol RT (8.31J/mol K)(308.15 K) Calculate the depth to which Avogadro’s number of table tennis balls would cover Earth. Each ball has a diameter of 3.75 cm. Assume the space between balls adds an extra 25.0% to their volume and assume they are not crushed by their own weight. Openstax College Physics Instructor Solutions Manual Chapter 13 Solution Let re = the radius of Earth, rp = the radius of the ping pong balls, and R = the radius to the top of the ping pong balls measured from Earth’s center. Then 4 4 3 3 V R 3 re N A rp (1.250) 3 3 R 3 re (1.250) N A rp 3 3 R (1.250) N A rp re 3 3 1 3 1 (1.250)(6. 02 10 23 )(1.875 10 2 m) 3 (6.376 10 6 m) 3 3 6.416 10 6 m h R re 4.04 10 4 m 40.4 km 37. (a) What is the gauge pressure in a 25.0C car tire containing 3.60 mol of gas in a 30.0 L volume? (b) What will its gauge pressure be if you add 1.00 L of gas originally at atmospheric pressure and 25.0C ? Assume the temperature returns to 25.0C and the volume remains constant. Solution (a) PV nRT nRT (3.60 mol)(0.082 1 L atm/mol K)(298.15 K ) P 2.937 atm V 30.0L Pg P 1.00 atm 1.937 atm 1.94 atm (b) PV nRT PV (1.00 atm )(1.00 L) Δn 0.0408 mol RT (0.0821 L atm/mol K )( 298.15K ) (n n) RT P V (3.60 mol 0.0408 mol )(0.0821 L atm/mol K )( 298.15 K ) 2.97 atm 30.0 L Pg P 1.00 atm 1.97 atm 38. (a) In the deep space between galaxies, the density of atoms is as low as 10 6 atoms/m 3 , and the temperature is a frigid 2.7 K. What is the pressure? (b) What volume (in m 3 ) is occupied by 1 mol of gas? (c) If this volume is a cube, what is the length of its sides in kilometers? Openstax College Physics Instructor Solutions Manual Chapter 13 Solution (a) PV NkT N 10 6 P kT (1.38 10 23 J/K)(2.7 K) 3 V 1m 3.73 10 17 N/m 2 3.7 10 17 N/m 2 3.7 10 17 Pa (b) PV nRT nRT (1.00 mol)(8.31 J/mol K)(2.7 K) V 6.02 1017 m 3 6.0 1017 m 3 17 2 P 3.73 10 N/m (c) L V 1/ 3 (6.02 1017 m3 )1/ 3 8.45 105 m 8.4 102 km 13.4 KINETIC THEORY: ATOMIC AND MOLECULAR EXPLANATION OF PRESSURE AND TEMPERATURE 39. Solution Some incandescent light bulbs are filled with argon gas. What is vrms for argon atoms near the filament, assuming their temperature is 2500 K? m rms 40. Solution 39.95 10 3 kg/mol 6.64 10 26 kg 23 6.02 10 / mol 3kT 3(1.38 10 23 J/K )( 2500 K ) 1.25 10 3 m/s 26 m 6.64 10 kg Average atomic and molecular speeds (vrms ) are large, even at low temperatures. What is vrms for helium atoms at 5.00 K, just one degree above helium’s liquefaction temperature? m rms 4.003 10 3 kg/mol 6.65 10 27 kg 23 6.02 10 / mol 3kT m 3(1.38 10 23 J/K )(5.00 K ) 176 m/s 6.65 10 27 kg 41. (a) What is the average kinetic energy in joules of hydrogen atoms on the 5500C surface of the Sun? (b) What is the average kinetic energy of helium atoms in a region 5 of the solar corona where the temperature is 6.00 10 K ? Solution (a) KE 3 kT 1.5(1.38 10 23 J/K )(5773.15 K) 1.195 10 19 J 1.20 10 19 J 2 3 (b) KE kT 1.5(1.38 10 23 J/K )(6.00 10 5 K) 1.24 10 17 J 2 Openstax College Physics 42. Instructor Solutions Manual Chapter 13 The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would oxygen molecules (molecular mass is equal to 32.0 g/mol) have an average velocity vrms equal to Earth’s escape velocity of 11.1 km/s? Solution (a) 11.2 km/s 1000 m 11200 m/s 1 km 60 s 60 minutes 11.2 km/s 40320 km/h 1 min 1 hour 11.2km/s (b) 3.20 10 2 kg/mol m 5.316 10 26 kg 23 6.02 10 / mol 3kT 3 1 2 rms kT mvrms m 2 2 2 m rms (5.316 10 26 kg)(1.11 10 4 m/s) 2 T 1.58 10 5 K 23 3k 3(1.38 10 J/K) 43. The escape velocity from the Moon is much smaller than from Earth and is only 2.38 km/s. At what temperature would hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity vrms equal to the Moon’s escape velocity? Solution 2.016 10 3 kg/mol 3.349 10 27 kg 23 6.02 10 / mol 3kT rms m 2 m rms (3.349 10 27 kg)(2.38 10 3 m/s) 2 T 458 K 3k 3(1.38 10 23 J/K) 44. Nuclear fusion, the energy source of the Sun, hydrogen bombs, and fusion reactors, occurs much more readily when the average kinetic energy of the atoms is high—that is, at high temperatures. Suppose you want the atoms in your fusion experiment to –14 have average kinetic energies of 6.40 10 J . What temperature is needed? Solution KE m 3 kT 2 2KE 2(6.40 10 14 J ) T 3.09 10 9 K 23 3k 3(1.38 10 J/K ) Openstax College Physics 45. Instructor Solutions Manual Chapter 13 Suppose that the average velocity (vrms ) of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol) in a flame is found to be 1.05 105 m/s . What temperature does this represent? Solution m rms 44.0 10 3 kg/mol 7.309 10 26 kg 6.02 10 23 / mol 3kT m T 2 m rms (7.309 10 26 kg)(1.05 105 m/s) 2 1.95 107 K 23 3k 3(1.38 10 J/K) 46. Hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity vrms equal to 193 m/s. What is the temperature? Solution 2.016 10 3 kg/mol 3.349 10 27 kg 23 6.02 10 / mol 3kT rms m 2 m rms (3.349 10 27 kg)(193 m/s) 2 T 3.01 K 3k 3(1.38 10 23 J/K) 47. Much of the gas near the Sun is atomic hydrogen. Its temperature would have to be 1.5 107 K for the average velocity vrms to equal the escape velocity from the Sun. m What is that velocity? Solution Since the sun consists of atomic hydrogen, not hydrogen molecules: 1.008 10 3 kg/mol m 1.674 10 27 kg 23 6.02 10 / mol rms 48. 3kT 3(1.38 10 23 J/K )(1.50 10 7 K ) 6.09 10 5 m/s 27 m 1.674 10 kg There are two important isotopes of uranium— 235 U and 238 U ; these isotopes are nearly identical chemically but have different atomic masses. Only 235 U is very useful in nuclear reactors. One of the techniques for separating them (gas diffusion) is based on the different average velocities vrms of uranium hexafluoride gas, UF6 . (a) The molecular masses for 235 U UF6 and 238 U UF6 are 349.0 g/mol and 352.0 g/mol, respectively. What is the ratio of their average velocities? (b) At what temperature would their average velocities differ by 1.00 m/s? (c) Do your answers in this problem imply that this technique may be difficult? Openstax College Physics Solution (a) Instructor Solutions Manual Chapter 13 (3kT ) / m235 v235 m238 m238 / N A m238 352.0 1.0043 1.004 v238 m235 m235 / N A m235 349.0 (3kT ) / m238 (b) v235 1.0043v238 v238 (0.0043 )v238 From this we see that the difference in their velocities is: v235 v238 (0.0043 )v238 and we want this to equal 1.00 m/s. (0.0043)v238 1.00 m/s 1.00 m/s (0.0043) 3kT m238 Solving for T : 2 2 vrms M 238 / N A v m T rms 238 2 3k (0.0043) 0.0043 3k 2 1 (1.00 m/s ) 3.52 10 kg/mol 764 K 2 (0.0043) 3(1.38 10 23 J/K)(6.02 10 23 /mol ) (c) This temperature is equivalent to 915F , which is hot but not impossible to achieve. Thus, this process is doable. At this temperature, however, there may be other considerations that make this process difficult. 13.6 HUMIDITY, EVAPORATION, AND BOILING 49. Dry air is 78.1% nitrogen. What is the partial pressure of nitrogen when the 5 2 atmospheric pressure is 1.01 10 N/m ? Solution (1.01105 N/m 2 )(0.781) 7.89 10 4 Pa 50. (a) What is the vapor pressure of water at 20.0C ? (b) What percentage of atmospheric pressure does this correspond to? (c) What percent of 20.0C air is water vapor if it has 100% relative humidity? (The density of dry air at 20.0C is 1.20 kg/m 3 .) Solution (a) Vapor Pressure for H 2O(20C) 2.33 103 N/m 2 2.33 103 Pa (b) Divide the vapor pressure by atmospheric pressure: 2.33 103 N/m 2 100% 2.30% 1.01 105 N/m 2 (c) The density of water in this air is equal to the saturation vapor density of water at this temperature, taken from Table 13.5. Dividing by the density of dry air, we can Openstax College Physics Instructor Solutions Manual get the percentage of water in the air: 51. Chapter 13 1.72 10 2 kg/m 3 100% 1.43% 1.20 kg/m 3 Pressure cookers increase cooking speed by raising the boiling temperature of water above its value at atmospheric pressure. (a) What pressure is necessary to raise the boiling point to 120.0C ? (b) What gauge pressure does this correspond to? Solution (a) Using Table 13.5: 1.99 105 N/m 2 1.99 105 Pa (b) 1.99 105 Pa 1.01105 Pa 9.8 10 4 Pa 0.97 atm 52. (a) At what temperature does water boil at an altitude of 1500 m (about 5000 ft) on a 4 2 day when atmospheric pressure is 8.59 10 N/m ? (b) What about at an altitude of 4 2 3000 m (about 10,000 ft) when atmospheric pressure is 7.00 10 N/m ? Solution (a) From Table 13.5: 95C (b) From Table 13.5: 90°C 53. What is the atmospheric pressure on top of Mt. Everest on a day when water boils there at a temperature of 70.0C? Solution From Table 13.5: 3.12 10 4 Pa 54. At a spot in the high Andes, water boils at 80.0C , greatly reducing the cooking speed of potatoes, for example. What is atmospheric pressure at this location? Solution From Table 13.5: 4.73 10 4 Pa 55. What is the relative humidity on a 25.0C day when the air contains 18.0 g/m 3 of water vapor? Solution Percent relative humidity 56. vapor density 100% saturation vapor density 18.0 g/m 3 100% 78.26% 78.3% 23.0 g/m 3 What is the density of water vapor in g/m 3 on a hot dry day in the desert when the temperature is 40.0C and the relative humidity is 6.00%? Openstax College Physics Instructor Solutions Manual Chapter 13 Solution percent relative humidity 57. A deep-sea diver should breathe a gas mixture that has the same oxygen partial pressure as at sea level, where dry air contains 20.9% oxygen and has a total pressure 5 2 of 1.01 10 N/m . (a) What is the partial pressure of oxygen at sea level? (b) If the diver breathes a gas mixture at a pressure of 2.00 106 N/m 2 , what percent oxygen should it be to have the same oxygen partial pressure as at sea level? vapor density 100% saturation vapor density percent relative humidity saturation vapor density vapor density 100% 3 (6.00%)(51.1 g/m ) 3.07 g/m 3 100% Solution (a) (1.013 105 N/m2 )(0.209) 2.117 104 N/m2 2.12 104 N/m2 2.12 104 Pa 6 2 4 2 (b) x(2.00 10 N/m ) 2.117 10 N/m x 2.117 10 4 N/m 2 1.059 10 2 1.06% 6 2 2.00 10 N/m 58. The vapor pressure of water at 40.0C is 7.34 103 N/m 2 . Using the ideal gas law, calculate the density of water vapor in g/m 3 that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature (51.1 g/m 3 ) . Solution n P 7.34 10 3 N/m 2 PV nRT 2.82 mol/m 3 V RT (8.31 J/mol K)(313.15 K) n M (2.82 mol/m 3 )(18.0 g/mol) 50.8 g/m 3 V 59. Air in human lungs has a temperature of 37.0C and a saturation vapor density of 44.0 g/m 3 . (a) If 2.00 L of air is exhaled and very dry air inhaled, what is the maximum loss of water vapor by the person? (b) Calculate the partial pressure of water vapor having this density, and compare it with the vapor pressure of 6.31 10 3 N/m2 . Solution (a) The maximum loss of water would occur if the exhaled air is completely saturated, and the inhaled air is devoid of water. The maximum water vapor loss is: (44.0 g/m 3 )(2.00 103 m3 ) 8.80 102 g (b) n M n , PV nRT P n RT RT V V M V M 3 44.0 g/m P (8.31 J/mol K)(310.15 K) 6.30 10 3 J/m 3 6.30 10 3 Pa 18.0 g/mol The two values are nearly identical. Openstax College Physics 60. Instructor Solutions Manual Chapter 13 If the relative humidity is 90.0% on a muggy summer morning when the temperature is 20.0C , what will it be later in the day when the temperature is 30.0C , assuming the water vapor density remains constant? Solution Let prh = percent relative humidity, vd = vapor density, and svd = saturation vapor density. vd vd prh 100%, prh 100% svd svd 17.2 g/m 3 svd 0.509 50.9% prh ' prh (0.900) 3 svd 30.4 g/m 61. Late on an autumn day, the relative humidity is 45.0% and the temperature is 20.0C . What will the relative humidity be that evening when the temperature has dropped to 10.0C , assuming constant water vapor density? Solution Let prh = percent relative humidity, vd = vapor density, and svd = saturation vapor density. vd vd prh 100%, prh 100% svd svd svd 17.2 g/m 3 prh = prh (0.450)( ) 82.3% svd 9.40 g/m 3 62. 4 2 Atmospheric pressure atop Mt. Everest is 3.30 10 N/m . (a) What is the partial pressure of oxygen there if it is 20.9% of the air? (b) What percent oxygen should a mountain climber breathe so that its partial pressure is the same as at sea level, where atmospheric pressure is 1.01 10 5 N/m 2 ? (c) One of the most severe problems for those climbing very high mountains is the extreme drying of breathing passages. Why does this drying occur? Openstax College Physics Instructor Solutions Manual Chapter 13 Solution (a) partial pressure O 2 (%O 2 )(atmosphe ric pressure) (0.209)(3. 30 10 4 N/m 2 ) 6.90 103 Pa (b) partial pressure at sea level (%O 2 )(atmosphe ric pressure) (0.209)(1. 013 105 N/m 2 ) 2.117 10 4 Pa Set that equal to the percent oxygen times the pressure at the top of Mt. Everest: %O 2 4 2 4 partial pressure at sea level (3.30 10 N/m ) 2.117 10 Pa 100% 4 2 2.117 10 N/m Thus, %O 2 100% 64.2% 3.30 10 4 N/m 2 (c) This drying process occurs because the partial pressure of water vapor at high altitudes is decreased substantially. The climbers breathe very dry air, which leads to a lot of moisture being lost due to evaporation. The breathing passages are therefore not getting the moisture they require from the air being breathed. 63. What is the dew point (the temperature at which 100% relative humidity would occur) on a day when relative humidity is 39.0% at a temperature of 20.0C ? Solution Let prh = percent relative humidity, vd = vapor density, and svd = saturation vapor density. vd (prh)(svd) (39.0%)(17.2 g/m 3 ) prh 100% vd 6.708 g/m 3 svd 100% 100% Now, assuming the relationship between temperature at saturation vapor pressure is T linear between 0C and 5C , we can use the fact that the value of m is svd constant between 0C and our temperature and between our temperature and 5C to interpolate, which yields: T( C ) svd( g/m 3 ) 0 4.84 4.7* 6.708 5 6.80 6.708 4.84 * 5C 4.77 C 6.80 4.84 Therefore, the Dew point 4.77C. 64. On a certain day, the temperature is 25.0C and the relative humidity is 90.0%. How many grams of water must condense out of each cubic meter of air if the temperature falls to 15.0C ? Such a drop in temperature can, thus, produce heavy dew or fog. Openstax College Physics Instructor Solutions Manual Chapter 13 Solution Let prh = percent relative humidity, vd = vapor density, and svd = saturation vapor density. vd (prh)(svd) (90.0%)( 23.0 g/m 3 ) prh 100% vd 20.7 g/m 3 svd 100% 100% 3 Water density leaving air = (20.7 g/m ) (12.8 g/m 3 ) 7.9 g/m 3 65. Integrated Concepts The boiling point of water increases with depth because pressure increases with depth. At what depth will fresh water have a boiling point of 150C , if the surface of the water is at sea level? Solution From Table 13.5: P 4.76 105 N/m 2 gh P A h 66. P PA 4.76 105 N/m 1.01105 N/m 38.3 m g 1000 kg/m 3 9.80 m/s 2 Integrated Concepts (a) At what depth in fresh water is the critical pressure of water reached, given that the surface is at sea level? (b) At what temperature will this water boil? (c) Is a significantly higher temperature needed to boil water at a greater depth? Solution (a) P gh P A P PA 22.12 10 6 N/m 2 1.0110 5 N/m 2 h 2246 m 2.25 103 m 3 3 2 g (1.00 10 kg/m )(9.80 m/s ) (b) At Tc 647.4 K from Table 13.5. (c) No. Water at a greater depth should boil for any temperature above the critical temperature. 67. Integrated Concepts To get an idea of the small effect that temperature has on Archimedes' principle, calculate the fraction of a copper block's weight that is supported by the buoyant force in 0C water and compare this fraction with the fraction supported in 95.0C water. Solution In 0C water: Let FB = the buoyant force, ww = the weight of water displaced, wC = the weight of the copper block, and VC = the volume of the copper block. FB ww wVC g FB wVC g w 1.00 10 3 kg/m 3 0.114 wC CVC g C 8.8 10 3 kg/m 3 In 95C water: Openstax College Physics ( Instructor Solutions Manual ' V g ' (1 C T ) FB ) w' C w' w wC CVC g C C (1 w T ) Chapter 13 1.00 10 3 kg/m 3 1 (5.110 5 / C)(95.0 C) 0.112 8.8 10 3 kg/m 3 1 (2.10 10 4 / C)(95.0 C) FB / wC 1.02 . As we can see, the buoyant force supports nearly the exact same ( FB / wC ) amount of force on the copper block in both circumstances. 68. Integrated Concepts If you want to cook in water at 150C , you need a pressure cooker that can withstand the necessary pressure. (a) What pressure is required for the boiling point of water to be this high? (b) If the lid of the pressure cooker is a disk 25.0 cm in diameter, what force must it be able to withstand at this pressure? Solution (a) From Table 13.5, vapor pressure 4.76 105 N/m 2 F . Here, we need to use Newton’s laws to balance forces. Assuming that we are A cooking at sea level, the forces on the lid will stem from the internal pressure, found in part (a), the ambient atmospheric pressure, and the forces holding the lid shut. Thus we have a “balance of pressures”: (b) P P 1 atm (4.76 10 5 Pa) = 0 P = 3.75 10 5 Pa net F = PA (3.75 10 5 Pa)( (0.125 m) 2 ) = 1.84 10 4 N 69. Unreasonable Results (a) How many moles per cubic meter of an ideal gas are there 14 2 at a pressure of 1.00 10 N/m and at 0C ? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? Solution 70. n P 1.00 1014 N/m 2 4.41 1010 mol/m 3 V RT (8.31 J/mol K)(273.15 K) (b) It is unreasonably large. (c) At high pressures such as these, the ideal gas law can no longer be applied. As a result, unreasonable answers come up when it is used. (a) PV nRT Unreasonable Results (a) An automobile mechanic claims that an aluminum rod fits loosely into its hole on an aluminum engine block because the engine is hot and the rod is cold. If the hole is 10.0% bigger in diameter than the 22.0C rod, at what temperature will the rod be the same size as the hole? (b) What is unreasonable about this temperature? (c) Which premise is responsible? Solution (a) Call the hole diameter d and the cold rod diameter d 0 . We have d 1.100d 0 . The Openstax College Physics Instructor Solutions Manual Chapter 13 area of the hole is A , and the cold rod area is A0 . The area of the rod at temperature T will be A A0 (2α) A0 T . Since the area is proportional to the diameter squared, we have: 2 2 2 d 0 (1.100) 2 d 0 2 2.5 10 5 / C d 0 (T 22C) 1.100 1 22C 4.2 10 3 C 5.0 10 / C (b) This temperature is unreasonably high—higher than the boiling point of aluminum. (c) The premise that the part can expand to 10% larger is unreasonable. This is much too great for almost any material. 2 T 71. 5 Unreasonable Results The temperature inside a supernova explosion is said to be 2.00 1013 K . (a) What would the average velocity vrms of hydrogen atoms be? (b) What is unreasonable about this velocity? (c) Which premise or assumption is responsible? Solution (a) m v rms 1.008 10 3 kg/mol 1.674 10 27 kg 23 6.02 10 /mol 3kT 3(1.38 10 23 J/K)(2.00 1013 K ) 7.03 10 8 m/s 27 m 1.674 10 kg (b) The velocity is too high—it is greater than the speed of light. (c) The assumption that hydrogen inside a supernova behaves as an ideal gas is unreasonable because of the great temperature and density in the core of a star. Furthermore, when a velocity close to the speed of light is obtained, classical physics must be replaced by relativity. 72. Unreasonable Results Suppose the relative humidity is 80% on a day when the temperature is 30.0C . (a) What will the relative humidity be if the air cools to 25.0C and the vapor density remains constant? (b) What is unreasonable about this result? (c) Which premise is responsible? Solution (a) Let prh = percent relative humidity, vd = vapor density, and svd = saturation vapor density. Openstax College Physics Instructor Solutions Manual Chapter 13 vd 100% svd 30.4g/m 3 vd svd 105.7% 106% prh 100% prh (0.800) 3 svd' svd 23.0 g/m prh (b) The answer is unreasonable because the maximum possible relative humidity is 100%. (c) The premise that the vapor density remains constant is unreasonable when the temperature drops below dew point, as evidenced by the 106% relative humidity. This file is copyright 2016, Rice University. All Rights Reserved.
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