Question

One mole of oxygen at 273 K and atmospheric pressure is adiabatically

compressed to 5 atm. Calculate the final temperature. Also calculate the work

done on the gas. Take gama = 1.4 and R = 8.31 J mol^-1 K^-1.

Answer 1

Solution:

Adiabatically compression of O₂:  P₁V₁¹·⁴ = P₂V₂¹·⁴

1) Combined gas law: $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

2) Then, $\dfrac{P_1^{1.4}V_1^{1.4}}{T_1^{1.4}}=\dfrac{P_2^{1.4}V_2^{1.4}}{T_2^{1.4}}$

$\dfrac{2}{1}: \dfrac{P_1^{0.4}}{T_1^{1.4}}=\dfrac{P_2^{0.4}}{T_2^{1.4}}$

Then, $\sqrt[1.4]{T₁¹·⁴ × (\frac{P₂}{P₁})⁰·⁴}$

Final temperature, $\sqrt[1.4]{273^{1.4}[\frac{5}{1}]^{0.4}}=432 K$

Work done one the gas = ΔPV = nRΔT = 1 × 8.31 × (432 - 273) W = 573 J