In This post you will get answer of Q.
A gas expands with temperature according to the relation, $V=k T^{2 / 3}$, where $k$ is a constant. Work done when the temperature changes by $60, K$ is $( R =$ universal gas constant. $)$
Solution:
Given, $V=k T^{2 / 3}$
From definition of work done,
$dW =P, dV=frac{R T}{V} d V $
$=frac{R T}{k T^{2 / 3}} d V,,,,,,,,,dots(i)$
Now, $V=k T^{2 / 3}$
Taking derivative on the both sides, we get
$d V=kappa frac{2}{3} T^{-1 / 3} d T,,,,,,,,dots(ii)$
Substituting the value of Eqs. (ii) in Eqs. (i) and taking integration on the both sides, we get
$W =frac{2}{3} R intlimits_{T_{1}}^{T_{2}} d T=frac{2}{3} Rleft(T_{2}-T_{1}right) $
$=frac{2}{3} R(60-0)=frac{2}{3} times R times 60$
$=40 ,R$
Solution:
Given, $V=k T^{2 / 3}$
From definition of work done,
$dW =P, dV=frac{R T}{V} d V $
$=frac{R T}{k T^{2 / 3}} d V,,,,,,,,,dots(i)$
Now, $V=k T^{2 / 3}$
Taking derivative on the both sides, we get
$d V=kappa frac{2}{3} T^{-1 / 3} d T,,,,,,,,dots(ii)$
Substituting the value of Eqs. (ii) in Eqs. (i) and taking integration on the both sides, we get
$W =frac{2}{3} R intlimits_{T_{1}}^{T_{2}} d T=frac{2}{3} Rleft(T_{2}-T_{1}right) $
$=frac{2}{3} R(60-0)=frac{2}{3} times R times 60$
$=40 ,R$