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
NEET 2017, Medium
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$V=\frac bT$
$VT=$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, $Q=n C \Delta T$
$=(1)\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T$
$=\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T$
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