MHT CET · Physics · Thermodynamics
A diatomic gas undergoes adiabatic change. Its pressure 'P' and temperature 'T' are related as \(\mathrm{P} \propto \mathrm{T}^{\mathrm{x}}\), where \(\mathrm{x}\) is
- A 3
- B 2.5
- C 3.5
- D 1.5
Answer & Solution
Correct Answer
(C) 3.5
Step-by-step Solution
Detailed explanation
(D)
\(\begin{array}{ll}\text { Adiabatic change } \rightarrow \mathrm{PV}^{\gamma}=\mathrm{k} & \mathrm{PV}=\mathrm{RT} \\ \mathrm{P} \frac{\mathrm{R}^{\gamma} \mathrm{T}^{\gamma}}{\mathrm{p}^{\gamma}}=\mathrm{k} & \mathrm{V}^{\gamma}=\left(\frac{\mathrm{RT}}{\mathrm{P}}\right)^{\gamma}\end{array}\)
\(\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma}=\mathrm{k}^{\prime}\)
\(\mathrm{P}^{1-\gamma}=\frac{\mathrm{k}^{\prime}}{\mathrm{T}^{\gamma}}\)
\(P=\frac{k^{n}}{T^{\gamma / 1-\gamma}}=k^{n} T\left(\frac{\gamma}{1-\gamma}\right)\)
\(\therefore \quad \mathrm{x}=\frac{\gamma}{\gamma-1}\)
For diatomic gas \(\gamma=1.4\)
\(\therefore \quad x=\frac{1.4}{0.4}=3.5\)
\(\begin{array}{ll}\text { Adiabatic change } \rightarrow \mathrm{PV}^{\gamma}=\mathrm{k} & \mathrm{PV}=\mathrm{RT} \\ \mathrm{P} \frac{\mathrm{R}^{\gamma} \mathrm{T}^{\gamma}}{\mathrm{p}^{\gamma}}=\mathrm{k} & \mathrm{V}^{\gamma}=\left(\frac{\mathrm{RT}}{\mathrm{P}}\right)^{\gamma}\end{array}\)
\(\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma}=\mathrm{k}^{\prime}\)
\(\mathrm{P}^{1-\gamma}=\frac{\mathrm{k}^{\prime}}{\mathrm{T}^{\gamma}}\)
\(P=\frac{k^{n}}{T^{\gamma / 1-\gamma}}=k^{n} T\left(\frac{\gamma}{1-\gamma}\right)\)
\(\therefore \quad \mathrm{x}=\frac{\gamma}{\gamma-1}\)
For diatomic gas \(\gamma=1.4\)
\(\therefore \quad x=\frac{1.4}{0.4}=3.5\)
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