Thermodynamics

JEE Main level, single correct, medium

146 of 190

Question 146 of 190

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

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Answer: (A) 12RΔT\frac{1}{2} \mathrm{R} \Delta \mathrm{T}

According to question VT=K\mathrm{VT}=\mathrm{K}

we also know that PV=nRTP V=n R T

⇒T=(PVnR)\Rightarrow \mathrm{T}=\left(\frac{\mathrm{PV}}{\mathrm{nR}}\right)

⇒V(PVnR)=k⇒PV2=K\Rightarrow \mathrm{V}\left(\frac{\mathrm{PV}}{\mathrm{nR}}\right)=\mathrm{k} \Rightarrow \mathrm{PV}^{2}=\mathrm{K}

∵C=R1−x+Cv\because \mathrm{C}=\frac{\mathrm{R}}{1-\mathrm{x}}+\mathrm{C}_{\mathrm{v}} (For polytropic process)

C=R1−2+3R2=R2\mathrm{C}=\frac{\mathrm{R}}{1-2}+\frac{3 \mathrm{R}}{2}=\frac{\mathrm{R}}{2}

∴ΔQ=nCΔT\therefore \Delta \mathrm{Q}=\mathrm{nC} \Delta \mathrm{T}

=R2×ΔT=\frac{R}{2} \times \Delta \mathrm{T} [here, n=1 mole]\left.\mathrm{n}=1 \mathrm{~mole}\right]

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