Question

Starting at temperature $300K,$ 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 $\left(i{n}^{\circ }K\right)$ is (to the nearest integer) ______.

Answer 1

Formula used: $T{V}^{\gamma -1}=constant$
Given, initial temperature of the gas, $T_1=300K$
And gas in compressed from volume $V_1$ to $V_2=\frac{V_1}{16}$
Let the temperature of the gas after compression be $T_2.$
For adiabatic process
$P{V}^{\gamma }=constant$
Also, $T{V}^{\gamma -1}=constant$
$300\times V_1^{\frac{7}{5}-1}=T_2{\left(\frac{V_1}{16}\right)}^{\frac{7}{5}-1}$
$300\times 2^{4\times \frac{2}{5}}=T_2$
Formula used: $PV=nRT$
Given, after compression gas expands isobarically to volume $2V_2.$
Let the final temperature of the gas be $T_f.$
From $PV=nRT,$

$V=\frac{nRT}{P}$
For isobaric process, $V=kT$
$V_2=k{T}_2\cdots \left(i\right)$
$2V_2=K{T}_f\cdots \left(ii\right)$
$\frac{1}{2}=\frac{T_2}{T_f}$
$\Rightarrow T_f=2T_2$
$T_f=2\times 300\times 2^{\frac{8}{5}}=1818.85\approx 1819$
Final answer: (1819.00)