MHT CET · Physics · Thermal Properties of Matter
In an adiabatic change, the pressure and temperature of a monoatomic gas are related with relation \(p \propto T^{C}\), where \(C\) is equal to
- A \(\frac{5}{4}\)
- B \(\frac{5}{3}\)
- C \(\frac{5}{2}\)
- D \(\frac{3}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
Key Idea Poisson's equation for adiabatic process is given by
\(
p V^{\gamma}=\text { constant. }
\)
For adiabatic process, Poisson's equation is given by
\(
p V^{\gamma}=\text { constant }
\)
Ideal gas relation is
\(
\begin{aligned}
& p V &=R T \\
\Rightarrow & V &=\frac{R T}{p}
\end{aligned}
\)
From Eqs. (i) and (ii), we get
\(
p\left(\frac{R T}{p}\right)^{\gamma}=\text { constant }
\)
\(
\Rightarrow \quad \frac{T^{\gamma}}{p^{\gamma-1}}=\text { constant }
\)
where \(\gamma\) is ratio of specific heats of the gas
Given, \(\quad p \propto T^{C} \quad\)... (iv)
On comparing with Eq. (iii), we have
\(
C=\frac{\gamma}{\gamma-1}
\)
For a monoatomic gas \(\gamma=\frac{5}{3}\)
We have
\(
C=\frac{\frac{5}{3}}{\frac{5}{3}-1}=\frac{5}{2}
\)
\(
p V^{\gamma}=\text { constant. }
\)
For adiabatic process, Poisson's equation is given by
\(
p V^{\gamma}=\text { constant }
\)
Ideal gas relation is
\(
\begin{aligned}
& p V &=R T \\
\Rightarrow & V &=\frac{R T}{p}
\end{aligned}
\)
From Eqs. (i) and (ii), we get
\(
p\left(\frac{R T}{p}\right)^{\gamma}=\text { constant }
\)
\(
\Rightarrow \quad \frac{T^{\gamma}}{p^{\gamma-1}}=\text { constant }
\)
where \(\gamma\) is ratio of specific heats of the gas
Given, \(\quad p \propto T^{C} \quad\)... (iv)
On comparing with Eq. (iii), we have
\(
C=\frac{\gamma}{\gamma-1}
\)
For a monoatomic gas \(\gamma=\frac{5}{3}\)
We have
\(
C=\frac{\frac{5}{3}}{\frac{5}{3}-1}=\frac{5}{2}
\)
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