Quiz 6 (08-09)
Chem 171 20 points
R = 0.0821 L x atm/mol x K = 8.3145 J/mol x K
Name_____________________/Character_________________
1)
When correcting the ideal gas law for actual gas behavior, a correction is made by assuming that the actual
volume for a gas to move in is ( smaller larger ) than for an ideal gas.
2)
The effusion rate of an unknown gas is measured and found to be 15.07 mL/min. Under identical conditions, the
effusion rate of O2 is 30.50 mL/min. If the choices are Cl2, Xe, Kr, H2, NO, CO2, what is the identity of the
unknown gas?
µx/µO2 =
𝑴 𝑶𝟐
𝑴 𝒙
𝒂𝒏𝒅
𝟏𝟓.𝟎𝟕
𝟑𝟎.𝟓𝟎
𝟑𝟐.𝟎𝟎
=
𝑴 𝒙
𝒂𝒏𝒅 𝟎. 𝟒𝟗𝟒 =
𝟑𝟐,𝟎𝟎
𝑴 𝒙
𝒂𝒏𝒅 𝟎. 𝟎. 𝟐𝟒 =
𝟑𝟐.𝟎𝟎
𝑴(𝒙)
M(x) = 131 which signifies Xe
A sample of nitrogen gas was collected over water at 20. C and a total pressure of 880 torr. The volume of the
gases was 350. mL. The vapor pressure of water at 20. C is 17.5 torr. What mass of nitrogen was collected?
3)
Path: Pressure of N2 to moles of N2 to mass of N2
Total pressure = PN2 + PH2O
Convert mL to L and C to K.
moles of N2
4)
=
𝑷𝑽
=
𝑹𝑻
880 torr = PN2 + 17.5 and PN2 = 862.5 torr
𝟏.𝟏𝟑 𝒂𝒕𝒎 𝟎.𝟑𝟓𝟎 𝑳
𝑳𝒂𝒕𝒎
𝒎𝒐𝒍𝑲
𝟎.𝟎𝟖𝟐𝟏
𝟐𝟗𝟑𝑲
= 𝟎. 𝟎𝟏𝟔 𝒎𝒐𝒍 𝒙
𝟐𝟖.𝟎 𝒈
𝒎𝒐𝒍
x 1 atm/760 torr = 1.13 atm
= 𝟎. 𝟒𝟔 𝒈
Which of the following is not consistent with the postulates of the kinetic molecular theory of gases?
a) pressure is due to collisions of gas particles with container walls
b) gas particles exert an attractive force on each other
c) the gas particle volume is insignificant
d) kinetic energy is directly proportional to temperature
5)
A glass container of volume 936 mL weighs 134.567 g when empty. It weighs 137.456 g when filled with gas to a
o
pressure of 735 torr at 31 C. What is the molar mass of this gas?
This can be solved two ways, for both solutions, a) pressure must be changed to atm b) temp must be in K
(conversion factor 1 atm/760 torr)
(K = C + 273)
Mass of chemical is obtained: mass of full container – mass of empty container = mass of chemical
137.456 g - 134.567 g = 2.889 g
Solution 1: Solve for moles and use total mass
𝑷𝑽
𝒏 = 𝑹𝑻 =
𝟎.𝟗𝟔𝟕 𝒂𝒕𝒎 (𝟎.𝟗𝟑𝟔 𝑳)
𝑳𝒂𝒕𝒎
𝒎𝒐𝒍𝑲
𝟎.𝟎𝟖𝟐𝟏
(𝟑𝟎𝟒)
= 𝟎. 𝟎𝟑𝟔𝟑 𝒎𝒐𝒍
Total mass/total moles = molar mass = 0.2.889g/0.0363 mol = 79.66 g/mol
Solution 2: Solve for density and plug values into M = dRT/P
density is mass/vol and is 0.2.889 g/0.936L = 3.07 g/L
M = (3.07 g/L)(0.0821 L x atm/mol x K)(304 K)/(0.967 atm)
6)
o
o
Consider three 1 L flasks.. Flask A contains NH3(g) at 25 C, flask B contains NO(g) at 25 C and flask C contains
o
N2(g) at 25 C.
a) In which flask do the molecules have the smallest average kinetic energy?
i) flask A
ii) flask B
iii) flask C
iv) all are the same (same temp)
b) In which flask do the molecules have the greatest velocity (rms)?
i) flask A
ii) flask B
iii) flask C
iv) all are the same (smallest)
7)
A rigid sealed 5.0 L container contains 2.0 g Cl2 and 7.00 g Ar.
a) What is the pressure of Cl2 gas when the temp is 293 K?
depends only on the moles of Cl2
2.0 g x mol/70.05 g = 0..0284 moles
𝑷=
Rearrange PV=nRT
𝒏𝑹𝑻
𝑽
=
𝑳𝒂𝒕𝒎
𝒎𝒐𝒍𝑲
𝟎.𝟎𝟐𝟖𝟒 𝒎𝒐𝒍 𝟎.𝟎𝟖𝟐𝟏
(𝟐𝟗𝟑 𝑲)
𝟓.𝟎 𝑳
= 𝟎. 𝟏𝟑𝟔 𝒂𝒕𝒎
b) What is the total pressure in the container?
This can be solved by adding individual pressures (Cl2 already determined in a and Ar solved the same) or by
adding moles and obtaining total pressure.
moles of Ar: 7.00 g Ar x mol/39.9 g = 0.175 mol, solving for P as above, gives a value of P ar = .842 atm
PTot = PH2 + PHe = 0.978 atm
𝑷=
OR
. 𝟏𝟕𝟓 𝒎𝒐𝒍 + 𝟎. 𝟎𝟐𝟖𝟒 𝒎𝒐𝒍
𝟎. 𝟎𝟖𝟐𝟏
𝟓. 𝟎 𝑳
𝑳𝒂𝒕𝒎
(𝟐𝟗𝟑𝑲)
𝒎𝒐𝒍𝑲
= 𝟎. 𝟗𝟕𝟖 𝒂𝒕𝒎
b) What is the mole fraction of Cl2?
Using moles: χH2 = 0..0284 mol/(0.0284 mol + 0.175 mol) = 0.139
Using pressure: χH2 = .136 atm/(.136 atm + .842 atm) = 0..139
c) If the temperature is increased to 313 K, which of the following will change?
i) the molar mass of Cl2
iii) the partial pressure of Cl2
8)
9)
ii) the volume of Cl
iv) the mole fraction of Cl2
A gas approaches ideal behavior under what conditions?
a) low T and high P
b) low T and low P
c) high T and low P
d) high T and high P
The reaction that forms propane is shown below. Consider what would occur if 0.300 mole of each
reactant is placed in a vessel and the reaction goes to completion.
3 CO (g) + 7 H2 (g) → C3H8 (g) + 3 H2O (g)
Set up a reaction table to help you see where the numbers come from.
a) How many moles of H2O would be produced?
Determine the limiting reagent: 0.3 mol CO x 3 mol H2O /3 mol CO = 0.3 mol H2O
0.3 mol H2 x 3 mol H2O/7 mol H2 = 0.13 mol H2O
H2 is limiting and 0.133 mol H2O is formed
b) How many moles of H2 would be left?
None, this is the limiting reagent.
c) How many moles of CO would be left?
0.3 mol H2 x 3 mol CO/7 mol H2 = 0.13 moles consumed
0.3 – 0.13 = 0.17 moles left
d) What would be the pressure due to C3H8 is the final conditions were 300 K and 0.2 atm?
Using H2 as the limiting reagent: 0.3 mol H2 x 1 mol C3H8/7 mol H2 = 0.043 mol C3H8
0.13 moles CO left; 0 moles H2 left; 0.133 moles H2O formed; 0.043 moles C3H8 formed
Total moles = 0.13 + 0.133 + 0.043 = 0.333 moles
Use mole fraction: mol C3H8/total moles x Ptot = 0.043/0.333 x 0.2 atm = 0.026 atm
e) What would is the total pressure if the vessel was 0.5 L and the temperature was 300 K?
Total pressure will depend on total number of moles: 0.333 from above
P = nRT/V = (0.333 mol)().0821 L x atm/mol x K)(300 K)/(0.5 L) = 16.40 atm