Question

During an experiment an ideal gas is found to obey the condition $\frac{P^2}{\rho }=\text{constant}$ [$\rho$ is the density of the gas]. The gas is initially at temperature $T$, pressure $P$ and density $\rho$. If the gas expands such that density changes to $\frac{\rho }{2}$ , then choose the correct answer(s).

• A. Pressure of the gas changes to $\sqrt{2}P$
• B. The temperature of the gas changes to $\sqrt{2}T$
• C. The $P-T$ Indicator diagram of the above process is a parabola.
• D. The $P-T$ Indicator diagram of the above process is a Rectangular hyperbola.

Answer 1

Answer: (B), (D)

The correct options are
B The temperature of the gas changes to $\sqrt{2}T$
D The $P-T$ Indicator diagram of the above process is a Rectangular hyperbola.

Given that, An ideal gas obeys a condition
$\frac{P^2}{\rho }=k......\left(1\right)$
$\Rightarrow \frac{P^{2}RT}{PM}=k$ $\because [\rho =\frac{M}{V}=\frac{M}{\frac{RT}{P}}]$
From this we can write that,
$PT=\frac{Mk}{R}=\text{constant}......\left(2\right)$
In comparision, the above formula looks similar to $xy=\text{constant}$
Which is an equation for Rectangular hyperbola.
$\therefore$ The $P-T$ graph gives a rectangular hyperbola for equation of state $\left(2\right)$.
From $\left(1\right)$ we can say that,
$\frac{P^2}{{\rho }_1}=\frac{P^{\prime 2}}{{\rho }_2}$
From the data given in the question,
$\frac{P^2}{\rho }=\frac{P^{\prime 2}}{\frac{\rho }{2}}\Rightarrow P^{{}^{\prime }}=\frac{P}{\sqrt{2}}\dots \dots \left(3\right)$
From$\left(2\right)$ we can write that
$\frac{P}{P^{\prime }}=\frac{T^{\prime }}{T}$
By using $\left(3\right)$ we get, $T^{{}^{\prime }}=\sqrt{2}T$
Thus, options (b) and (d) are the correct answers.