Question

Two moles of an ideal mono atomic gas undergoes compression from A to B as shown in the figure:


If for the process AB, $P{T}^2$ = constant
Then the work done (in kcal) on the system for the process AB is

Answer 1

From the graph,
$T_1=1000K$
$T_2=500K$
Given for the process $P{T}^2=C_1$ and we know PV = nRT
So $P{\left(\frac{PV}{nR}\right)}^2=C_1\Rightarrow P^3{V}^2=C$
$\Rightarrow P={\left(\frac{C}{V^2}\right)}^{\frac{1}{3}}$
$W=-{\int }_{V_1}^{V_2}PdV=-{\int }_{V_1}^{V_2}{C}^{\frac{1}{3}}\frac{dV}{V^{\frac{2}{3}}}$
$=-C^{\frac{1}{3}}3(V_2^{\frac{1}{3}}-V_1^{\frac{1}{3}})$
$=-3(P_2{V}_2-P_1{V}_1)$
$=3nR(T_1-T_2)$
$=3\times 2\times 2\times 500$
$=6000Cal=6kCal$