In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the…

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $\mathrm{VT}=\mathrm{K},$ where $\mathrm{K}$ is a constant. In this process the temperature of the gas is increased by $\Delta \mathrm{T}$. The amount of heat absorbed by gas is (R is gas constant):
  1. $\frac{1}{2} \mathrm{R} \Delta \mathrm{T}$
  2. $\frac{1}{2} \mathrm{KR} \Delta \mathrm{T}$
  3. $\frac{3}{2} \mathrm{R} \Delta \mathrm{T}$
  4. $\frac{2 \mathrm{~K}}{3} \Delta \mathrm{T}$

Solution

According to question $\mathrm{VT}=\mathrm{K}$ we also know that $P V=n R T$ $\Rightarrow \mathrm{T}=\left(\frac{\mathrm{PV}}{\mathrm{nR}}\right)$ $\Rightarrow \mathrm{V}\left(\frac{\mathrm{PV}}{\mathrm{nR}}\right)=\mathrm{k} \Rightarrow \mathrm{PV}^{2}=\mathrm{K}$ $\because \mathrm{C}=\frac{\mathrm{R}}{1-\mathrm{x}}+\mathrm{C}_{\mathrm{v}}$ (For polytropic process) $\mathrm{C}=\frac{\mathrm{R}}{1-2}+\frac{3 \mathrm{R}}{2}=\frac{\mathrm{R}}{2}$ $\therefore \Delta \mathrm{Q}=\mathrm{nC} \Delta \mathrm{T}$ $=\frac{R}{2} \times \Delta \mathrm{T}$ [here, $\left.\mathrm{n}=1 \mathrm{~mole}\right]$

Asked in: JEE Main 2019 (11 Jan Shift 2)

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