$(a)\;\large\frac{6}{13}\;\large\frac{p_{0}V_{0}}{nR}\qquad(b)\;-\large\frac{6}{13}\;\large\frac{p_{0}V_{0}}{nR}\qquad(c)\;\large\frac{1}{13}\;\large\frac{p_{0}V_{0}}{nR}\qquad(d)\;-\large\frac{1}{13}\;\large\frac{p_{0}V_{0}}{nR}$

Answer : $\;-\large\frac{6}{13}\;\large\frac{p_{0}V_{0}}{nR}$

Explanation :

Let $\;T_{1}\;$ be the initial temperature

$T_{1}=(\large\frac{p_{0}}{1+1})\;(\large\frac{V_{0}}{nR})$

Let $\;T_{2}\;$ be the final temperature

Therefore , $\;T_{2}=(\large\frac{p_{0}}{1+25})\;(\large\frac{V_{0}}{nR})$

$\bigtriangleup T=T_{2}-T_{1}\large\frac{p_{0}V_{0}}{nR}\;(\large\frac{1}{26}-\large\frac{1}{2})=-\large\frac{6p_{0}V_{0}}{13nR}$

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