A gas of mass ' $\mathrm{m}$ ' and molecular weight ' $\mathrm{M}$ ' is flowing in an insulated tube with a…

A gas of mass ' $\mathrm{m}$ ' and molecular weight ' $\mathrm{M}$ ' is flowing in an insulated tube with a velocity ' $2 \mathrm{~V}$ '. If the flow of the gas is suddenly stopped and all the kinetic energy is utilized to compress the gas, the increases in the temperature of the gas is ( $\gamma$ is ratio of specific heats, $\mathrm{R}$ is universal gas constant)
  1. $\frac{2 \mathrm{MV}^2(\gamma-1)}{\mathrm{R}}$
  2. $\frac{\mathrm{mV}^2(\gamma-1)}{2 \mathrm{MR}}$
  3. $\frac{\mathrm{mV}^2 \gamma}{2 \mathrm{R}}$
  4. $\frac{\mathrm{MV}^2 \gamma}{2 \mathrm{R}}$

Solution

Since the gas flow is suddenly stopped. We will Consider it to be an adiabatic process. Work done in a adiabatic process $ \mathrm{W}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\gamma-1}=\frac{\mathrm{mR} \Delta \mathrm{T}}{\mathrm{M}(\gamma-1)} $ Energy available after the gas flow suddenly stopped $ \begin{aligned} & =\frac{1}{2} \mathrm{~m}(2 \mathrm{~V})^2 \\ & \frac{1}{2} \mathrm{~m} \times 4 \mathrm{~V}^2=\frac{\mathrm{mR} \Delta \mathrm{T}}{\mathrm{M}(\gamma-1)} \\ & \Rightarrow \Delta \mathrm{T}=\frac{2 \mathrm{MV}^2(\gamma-1)}{\mathrm{R}} \end{aligned} $

Asked in: AP EAMCET 2023 (19 May Shift 1)

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