MHT CET · Physics · Thermodynamics
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
- A \(\frac{7}{2}\)
- B \(\frac{7}{3}\)
- C 5
- D 5
Answer & Solution
Correct Answer
(A) \(\frac{7}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{P} \propto \mathrm{~T}^{\mathrm{K}} \\
& \mathrm{PT}^{-\mathrm{K}}=\text { constant }
\end{aligned}\)
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),
\(\therefore \quad \frac{\gamma}{1-\gamma}=\frac{1.4}{1-1.4}=\frac{-1.4}{0.4}=\frac{-14}{4}=\frac{-7}{2}\)
\(\begin{aligned} & \therefore \quad \frac{-7}{2}=-K \\ & \therefore \quad K=\frac{7}{2}\end{aligned}\)
& \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),
\(\therefore \quad \frac{\gamma}{1-\gamma}=\frac{1.4}{1-1.4}=\frac{-1.4}{0.4}=\frac{-14}{4}=\frac{-7}{2}\)
\(\begin{aligned} & \therefore \quad \frac{-7}{2}=-K \\ & \therefore \quad K=\frac{7}{2}\end{aligned}\)
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