In an adiabatic process for an ideal gas, the relation between the universal gas constant ' $R$ ' and…

In an adiabatic process for an ideal gas, the relation between the universal gas constant ' $R$ ' and specific heat at constant volume ' $\mathrm{C}_{\mathrm{v}}$ ' is $R=0 \cdot 4 C_v$. The pressure ' $P$ ' of the gas is proportional to the temperature ' T ', of the gas as $\mathrm{T}^k$. The value of constant ' K ' is
  1. $\frac{7}{2}$
  2. $\frac{7}{3}$
  3. 5
  4. 5

Solution

$\begin{aligned} & \mathrm{P} \propto \mathrm{~T}^{\mathrm{K}} \\ & \mathrm{PT}^{-\mathrm{K}}=\text { constant } \end{aligned}$
For an adiabatic process, $\begin{array}{ll} & \mathrm{PT}^{\frac{\gamma}{1-\gamma}}=\text { constant } \\ \therefore \quad & \frac{\gamma}{1-\gamma}=-\mathrm{K}...(i) \\ & \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R} \\ \therefore \quad & \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+0.4 \mathrm{C}_{\mathrm{v}} \\ \mathrm{C}_{\mathrm{p}}=1.4 \mathrm{C}_{\mathrm{v}} \\ & \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=1.4...(ii) \end{array}$
From(i) and (ii), $\therefore \quad \frac{\gamma}{1-\gamma}=\frac{1.4}{1-1.4}=\frac{-1.4}{0.4}=\frac{-14}{4}=\frac{-7}{2}$ $\begin{aligned} & \therefore \quad \frac{-7}{2}=-K \\ & \therefore \quad K=\frac{7}{2}\end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 1)

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