Question

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand reversibly and adiabatically until its volume becomes 5.66 V, while its temperature decreases to $\frac{T}{2}$. Work done by the gas during expansion is given by $W=-XPV$. Find the value of $X$. (given $(5.66{)}^{0.4}=2$) Report the answer upto two decimals

Answer 1

For an adiabatic process,
$T{V}^{\gamma -1}=\text{constant}$
$\therefore T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}=\frac{T_1}{2}(5.66V_1{)}^{\gamma -1}$
Hence, $2=(5.66{)}^{\gamma -1}$
or $\text{log}2=(\gamma -1)\text{log}5.66$
$\therefore \gamma =1.4$
The gas is, therefore a diatomic gas and have five degrees of freedom. The work done by a gas during an adiabatic process is
$W=\frac{P_2{V}_2-P_1{V}_1}{1-\gamma }$
Since $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }=P_2(5.66V_1{)}^{\gamma }$
$P_2=\frac{P_1}{(5.66{)}^{\gamma }}$
$\therefore W=[\frac{P_1\left(5.66V_1\right)}{(5.66{)}^{\gamma }}-P_1{V}_1]\frac{1}{0.4}=-1.25P_1{V}_1$
or $W=-1.25PV$
Hence; $X=1.25$