> 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. $)$ – LIVE ANSWER TODAY

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. $)$

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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) $

$=frac{2}{3} R(60-0)=frac{2}{3} times R times 60$

$=40 ,R$

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