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Heat and Thermodynamics question

2011 · Shift 0 · Q69
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Heat and Thermodynamics question

2011 · Shift 0 · Q69

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A thermally insulated vessel contains an ideal gas of molecular mass MMM and ratio of specific heats γ.\gamma .γ. It is moving with speed vvv and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by:
  1. A
    (γ−1)2γRMv2K{{\left( {\gamma - 1} \right)} \over {2\gamma R}}M{v^2}K2γR(γ−1)​Mv2K
  2. B
    γM2v2RK{{\gamma {M^2}v} \over {2R}}K2RγM2v​K
  3. C
    (γ−1)2RMv2K{{\left( {\gamma - 1} \right)} \over {2R}}M{v^2}K2R(γ−1)​Mv2K
  4. D
    (γ−1)2(γ+1)RMv2K{{\left( {\gamma - 1} \right)} \over {2\left( {\gamma + 1} \right)R}}M{v^2}K2(γ+1)R(γ−1)​Mv2K
View written solutionFree

Correct answer: C

  1. Key idea

When the thermally insulated vessel is suddenly brought to rest, no heat is exchanged with the surroundings: Q=0Q=0Q=0

The macroscopic kinetic energy of the gas-vessel system gets converted into the internal energy of the gas.

  1. Loss of kinetic energy of the gas

If the vessel and gas were moving together initially with speed vvv, then for nnn moles of gas, total mass of gas is m=nMm=nMm=nM where MMM is the molar mass.

So the initial translational kinetic energy of the gas is K=12mv2=12nMv2K=\frac{1}{2}mv^2=\frac{1}{2}nMv^2K=21​mv2=21​nMv2

When the vessel is brought to rest, this kinetic energy is lost and appears as increase in internal energy: ΔU=12nMv2\Delta U=\frac{1}{2}nMv^2ΔU=21​nMv2

  1. Relate internal energy change to temperature rise

For an ideal gas, ΔU=nCVΔT\Delta U=nC_V\Delta TΔU=nCV​ΔT

Also, γ=CPCV,CP−CV=R\gamma=\frac{C_P}{C_V}, \qquad C_P-C_V=Rγ=CV​CP​​,CP​−CV​=R

Hence, CV=Rγ−1C_V=\frac{R}{\gamma-1}CV​=γ−1R​

Therefore, nCVΔT=12nMv2nC_V\Delta T=\frac{1}{2}nMv^2nCV​ΔT=21​nMv2

Substitute CVC_VCV​: nRγ−1ΔT=12nMv2n\frac{R}{\gamma-1}\Delta T=\frac{1}{2}nMv^2nγ−1R​ΔT=21​nMv2

Cancel nnn: Rγ−1ΔT=12Mv2\frac{R}{\gamma-1}\Delta T=\frac{1}{2}Mv^2γ−1R​ΔT=21​Mv2

So, ΔT=γ−12RMv2\Delta T=\frac{\gamma-1}{2R}Mv^2ΔT=2Rγ−1​Mv2

  1. Match with options

This matches: (γ−1)2RMv2 K\boxed{\frac{(\gamma-1)}{2R}Mv^2\text{ K}}2R(γ−1)​Mv2 K​

So the correct option is C.

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