Question

Two moles of a gas $(\gamma =5/3)$ are initially at temperature ${27}^{\circ }C$ and occupy a volume of $20L$. The gas is first expanded at constant pressure until the volume is doubled. Then it is subjected to an adiabatic change until the temperature returns to its initial value. The work done by the gas in $J$ is $1.2479\times {10}^x$. Find $x$.

Answer 1

Let the initial pressure be $P_o$.

In the first expansion,

Work done=$P_o(V_1-V_o)=P_o{V}_o=nR{T}_o$

$V\propto T$

$⟹T_1=\frac{40}{20}{T}_o=2T_o$

In the second process, the final temperature is reached from $2T_o$ to $T_o$.

For this adiabatic process, $T{V}^{\gamma -1}=constant$

Hence, $\left(600\right)(40{)}^{2/3}=(300\left)\right(V_2{)}^{2/3}$

$⟹V_2=40\times 2^{3/2}L$

Also $P{V}^{\gamma }=constant$

$⟹P_o(40{)}^{5/3}=P_2(40\times 2^{3/2}{)}^{5/3}$

$⟹P_2=P_o\times 2^{-5/2}$

Work done =$\frac{P_2{V}_2-P_1{V}_1}{1-\gamma }$

$=\frac{3}{2}nR{T}_o$

Hence total work done=$\frac{5}{2}nR{T}_o=12479J$