Question

1 mole of $C{O}_2$ gas at 300 K is expanded under a reversible adiabatic condition such that its volume becomes 27 times.
(a) What is the final temperature? (b) What is the work done?
(Given $\gamma$ = 1.33 and $C_v$ = 25.08 J ${mol}^{-1}{K}^{-1}$ for ${CO}_2$)

Answer 1

Given that $\gamma =1.33$ and $C_v=25.08Jmo{l}^{-1}{K}^{-1}$ for $C{O}_2$

$\left(a\right)$ For adiabatic condition $T_2/T_1={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$

$T_2=\left[{\left(\frac{1}{27}\right)}^{1.33-1}\right]\times 300=100K$

Thus $T_2>T_1$ $,$ hence cooling takes place due to the expansion under adiabatic condition$.$

$\left(b\right)$ $△E$

$q=0$ $($ for adiabatic condition$)$ therefore

$△E=w$

Work is done of the cost of internal energy hence internal energy decreases$.$

$w=-△E=-Cv\left(T_2-T_1\right)=-25.03\times \left(100-300\right)$

$=-5016J$