Question

Starting at temperature $300\text{K}$, one mole of an ideal diatomic gas $(\gamma =1.4)$ is first compressed adiabatically from volume $V_1$ to $V_2=\frac{V_1}{16}$. It is then allowed to expand isobarically to volume $2V_2$. If all the processes are the quasi-static, then the final temperature of the gas (in $K$) is (to the nearest integer)

Answer 1

For an adiabatic process,

$T{V}^{\gamma -1}=\text{constant}$
$\therefore T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$\Rightarrow T_2=T_1\times {\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$

$=\left(300\right)\times {\left(\frac{V_1}{\frac{V_1}{16}}\right)}^{1.4-1}$

$\Rightarrow T_2=300\times (16{)}^{0.4}=300\times (2^4{)}^{\frac{2}{5}}=300\times 2^{\frac{8}{5}}$
Ideal gas equation, $PV=nRT$
$\therefore V=\frac{nRT}{P}$

$\Rightarrow V=kT$ (since pressure is constant for isobaric process)

So, during isobaric process
$V_2=k{T}_2....\left(i\right)$

$2V_2=k{T}_f....\left(ii\right)$

Dividing $\left(i\right)$ by $\left(ii\right)$

$\frac{1}{2}=\frac{T_2}{T_f}$

$T_f=2T_2=300\times 2\times 2^{\frac{8}{5}}\approx 1818K$