One mole of an ideal gas γ = 5/3) expands adiabatically and quasi-statically from a pressure P₁ = 6 atm and a temperature of 50^oC to a pressure P₂ = 4 atm. How much work is done by the gas during this process? R = 8.314 J/mol · K = 8.206 L · atm/mol · K.

Question

Quizplus · Accepted Answer

We can use the adiabatic work formula: $W=\frac{\gamma}{\gamma - 1}P_1V_1\left(1-\left(\frac{P_2}{P_1}ight)^{\frac{\gamma-1}{\gamma}}ight)$
Using the ideal gas law, we can find that $V_1=\frac{nRT_1}{P_1}=9.12$ L and $V_2=\frac{nRT_2}{P_2}=12.18$ L. 
Substituting the given values into the adiabatic work formula, we get 
$W=\frac{5}{3-1}(6	ext{ atm}\cdot9.12	ext{ L})\left(1-\left(\frac{4}{6}ight)^\frac{2}{5}ight)=56.2$ kJ.

One Mole of an Ideal Gas γ = 5/3) Expands