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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
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Verified Answer
The correct answer is:
$\frac{5}{2}$
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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