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The volume of $1\; mole$ of an ideal gas with the adiabatic exponent $\gamma$ is changed according to the relation $V=\frac bT$ where $b =$ constant. The amount of heat absorbed by the gas in the process if the temperature is increased by $\triangle T$ will be
$\frac{R}{{\gamma - 1}} \Delta T$
$\left( {\frac{{2 - \gamma }}{{\gamma - 1}}} \right)R \Delta T$
$\;\frac{{R \Delta T}}{{\gamma - 1}}$
$\left( {\frac{{1 - \gamma }}{{\gamma + 1}}} \right)R \Delta T$
Solution
$V=\frac bT$
$VT=$constant
$V(p V)=$ constant
$\therefore p V^{2}=$ constant
In the process $p V^{x}=$ constant, molar heat capacity is
$C=\frac{R}{\gamma-1}+\frac{R}{1-x}$
Here, $x=2$
$\therefore C=\frac{R}{\gamma-1}+\frac{R}{1-2}=\left(\frac{2-\gamma}{\gamma-1}\right) R$
Now, $Q=n C \Delta T$
$=(1)\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T$
$=\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T$
