Thermodynamics

JEE Main level, single correct, medium

40 of 190

Question 40 of 190

A thermally insulated vessel contains an ideal gas of molecular mass MM and ratio of specific heats 1.41.4. Vessel is moving with speed vv and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by :
(R=R= universal gas constant)

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Answer: (B) Mv25R\dfrac{M{v}^{2}}{5R}

KE of the vessel =12mv2=\dfrac{1}{2}m{v}^{2}

When the vessel is suddenly stopped, the ordered motion of the gas molecules is converted into disordered motion of the molecules increasing thereby the internal energy of the gas. Thus,

∆U=nCv∆T=12mv2=12(nM)v2∆U=n{C}_{v}∆T=\dfrac{1}{2}m{v}^{2}=\dfrac{1}{2}\left(nM\right){v}^{2}

Where nn is number of moles of the gas in the vessel and MM is molecular weight of the gas.

∴∆T=Mv22Cv∴∆T=\dfrac{M{v}^{2}}{2{C}_{v}}

As Cv=Rγ−1{C}_{v}=\dfrac{R}{\gamma -1}

∴∆T=Mv2(γ−1)2R=Mv2(1.4−1)2R=Mv25R∴∆T=\dfrac{M{v}^{2}(\gamma -1)}{2R}=\dfrac{M{v}^{2}(1.4-1)}{2R}=\dfrac{M{v}^{2}}{5R}

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