Answer on Question#39781, Physics, Molecular physics
One mole of oxygen at STP is adiabatically compressed to 5 atm. Calculate the final temperature. Also,
calculate the work done on the gas. Take g = 1.4 and R = 8.31 J mol^-1 K^-1.
Solution
The equation of adiabatic process for an ideal gas is
𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡,
where 𝑃 is pressure, 𝑉 is volume, and 𝛾 = 1.4 - the adiabatic index of gas.
Also we know the state equation for an ideal gas:
𝑃𝑉 = 𝜐𝑅𝑇,
where 𝜐 is the amount of substance of gas, 𝑇 is the temperature of the gas and 𝑅 is the universal gas
constant.
So
𝑉=
𝜐𝑅𝑇
𝜐𝑅𝑇 𝛾
) = 𝑐𝑜𝑛𝑠𝑡 → 𝑃1−𝛾 ∙ 𝑇 𝛾 = 𝑐𝑜𝑛𝑠𝑡.
𝑎𝑛𝑑 𝑃 (
𝑃
𝑃
Hence
1−𝛾
𝑃
1−𝛾
𝛾
1−𝛾
𝛾
𝑃1 ∙ 𝑇1 = 𝑃2 ∙ 𝑇2 → 𝑇2 = ( 1
1−𝛾
𝑃2
1
𝛾 𝛾
∙ 𝑇1
) .
STP is (𝑃1 = 100 𝑘𝑃𝑎, 𝑇1 = 273 𝐾), 𝑃2 = 507 𝑘𝑃𝑎.
1
(100 ∙ 103 )1−1.4 ∙ 2731.4 1.4
𝑇2 = (
) = 434 𝐾.
(507 ∙ 103 )1−1.4
The work done on the gas in adiabatic process is
𝛾−1
𝛾
𝛼
𝑃2
𝑊 = 𝜐𝑅𝑇1 (( )
2
𝑃1
where 𝛼 =
2
𝛾−1
− 1),
= 5.
1.4−1
1.4
5
507 ∙ 103
((
𝑊 = ∙ 1 ∙ 8.31 ∙ 273
)
2
100 ∙ 103
Answer: 𝟒𝟑𝟒 𝑲; 𝟑𝟑𝟏 𝑱.
− 1) = 331 𝐽.