Question

During an experiment, an ideal gas is found to obey a condition [\rho=density of gas]. The gas is initially at temperature T , pressure P and density $\rho$. The gas expands such that density changes to $\dfrac{\rho }{2}$.
A. The pressure of the gas changes to $\sqrt 2 {P}$ $\dfrac{{{P^2}}}{\rho } = {\text{constant}}$
B. The temperature of the gas changes to $\sqrt 2 T$
C. The graph of the above process on the P-T diagram is parabola
D. The graph of the above process on the P-T diagram is hyperbola

Answer 1

Using the condition that the ideal gas obeys, calculate the changed pressure of the gas. Then using the ideal gas equation for density of the gas, calculate the changed temperature of the gas. Using the relation between the pressure and temperature of the gas, guess what should be the shape of the P-T graph of the given process.

Formula used:
The ideal gas equation for the density $\rho$ of the gas is
$\rho = \dfrac{{PM}}{{RT}}$ …… (1)
Here, $P$ is pressure of the gas, $M$ is mass of the gas, $R$ is gas constant and $T$ is temperature of the gas.

Complete step by step answer:
We have given that the initial temperature, pressure and density of the gas are $T$ , $P$ and $\rho$ respectively.
The changed density of the gas is $\dfrac{\rho }{2}$.
${\rho _f} = \dfrac{\rho }{2}$
We have given that the ideal gas used in an experiment obeys the relation
$\dfrac{{{P^2}}}{\rho } = {\text{constant}}$
Here, $P$ is pressure of the gas and $\rho$ is density of the gas.
$\Rightarrow P \propto \sqrt \rho$
Write the above relation for initial and final values of pressure and density.
$\Rightarrow \dfrac{{{P_i}}}{{{P_f}}} = \sqrt {\dfrac{{{\rho _i}}}{{{\rho _f}}}}$
The suffix I denotes initial and f denotes final values.
Substitute $P$ for ${P_i}$, $\rho$ for ${\rho _i}$ and $\dfrac{\rho }{2}$ for ${\rho _f}$ in the above equation.
$\Rightarrow \dfrac{P}{{{P_f}}} = \sqrt {\dfrac{\rho }{{\dfrac{\rho }{2}}}}$
$\Rightarrow {P_f} = \dfrac{P}{{\sqrt 2 }}$
The pressure of the gas changes to $\dfrac{P}{{\sqrt 2 }}$ and not to $P\sqrt 2$.
Hence, the option A is incorrect.
Substitute ${P^2}$ for $\rho$ inn equation (1).
${P^2} = \dfrac{{PM}}{{RT}}$
$\Rightarrow P = \dfrac{M}{{RT}}$
As the mass of the gas and gas constant in the above equation are constant, we can write from the above equation
$P \propto \dfrac{1}{T}$ …… (2)
Write the above relation for initial and final pressure and temperature.
$\dfrac{{{P_i}}}{{{P_f}}} = \dfrac{{{T_f}}}{{{T_i}}}$
Substitute $P$ for ${P_i}$, $\dfrac{P}{{\sqrt 2 }}$ for ${P_f}$ and $T$ for ${T_i}$ in the above equation.
$\dfrac{P}{{\dfrac{P}{{\sqrt 2 }}}} = \dfrac{{{T_f}}}{T}$
$\Rightarrow {T_f} = \sqrt 2 T$
Therefore, the changed temperature of the gas becomes $\sqrt 2 T$.
Hence, the option B is correct.
From equation (2), we can say the pressure and temperature of the ideal gas are inversely proportional to each other.
The pressure-temperature graph of the above process cannot be a parabola.
Hence, the option C is incorrect.
The pressure-temperature graph of the above mentioned process should be a hyperbola as the pressure and temperature are inversely proportional.
Hence, the option D is correct.

Hence, the correct options are B and D.

Note:
We have not actually drawn the pressure-temperature graph of the given process and just directly given the nature of the graph from the relation of pressure and temperature of the gas. One can check the hyperbolic nature of the gas by drawing the actual pressure-temperature graph of the process.