Question

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

Answer 1

Given:-

$\gamma =1.33$

$C_V=25.08J{mol}^{-1}{K}^{-1}$

(a) For adiabatic condition-

$T{V}^{\gamma -1}=\text{constant}$

$\frac{T_1}{T_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}$

Given:-

$T_1=300K$

$V_1=V$

$V_2=27V$

$T_2=T\left(\text{Say}\right)=?$

$\frac{300}{T}={\left(\frac{27V}{V}\right)}^{\frac{4}{3}-1}$

$\Rightarrow T=100K$

Hence final teperature will be $100K$.

(b) As we know that,

$\Delta E=q+w$

$\because q=0\left(\text{For adiabatic}\right)$

$\therefore w=\Delta E$

Work is done at the cost of internal energy; if gas expands, we have negative sign hence internal energy decreases.

$\therefore w=-\Delta E=-n{C}_{V}dT$

$\Rightarrow w=-1\times 25.08\times (100-300)=5016J$

Hence work done will be $5016J$.