Question

An ideal gas of volume V and pressure expands isothermally to volume 16V and then compressed adiabatically to volume V. The final pressure of a gas is: [$\gamma = 1.5$]
A.P
B.3P
C.4P
D.6P

Answer 1

Hint:Ideal gas is defined as that gas in which all collisions between molecules are perfectly elastic and there is no interactive force of attraction is observed. The equation for Ideal gas is given as:
$PV = nRT$
Here, Pressure is determined by ‘P’, Volume is represented by ’V’ and ‘n’ is the number of moles and ‘R’ is the gas constant and ‘T’ is represented by Temperature.

Complete step by step answer:
We know in case of isothermal process temperature is constant and according to Boyle's law at a constant temperature, the volume of a given amount of a gas is inversely proportional to the pressure of that gas. The relation between Pressure and Volume can be mathematically written as:
$P \propto \dfrac{1}{V}$
$ \Rightarrow {\text{PV}} = {\text{Constant}}$
Or
${P_1}{V_1} = {P_2}{V_2}$ (1)
Suppose ${P_1} = P$ and ${V_1} = V$, Final volume ${V_2}$ is given as $16V$. We need to find the final pressure ${P_2}$. On putting the value in equation 1, we get:
$P \times V = {P_2} \times 16V$
$ \Rightarrow {P_2} = \dfrac{{P \times V}}{{16V}} = \dfrac{P}{{16}}$
As we know for the adiabatic process the equation is written as:
$P{V^\gamma } = {\text{constant}}$
Or we can write:
${P_2}{({V_2})^\gamma } = {P_3}{({V_3})^\gamma }$ (2)
On putting the value of ${P_2},{V_2}$ and $\gamma $ in equation (2) we get:
$\dfrac{P}{{16}} \times {(16V)^{1.5}} = {P_3}{(V)^{1.5}}$
$ \Rightarrow \dfrac{P}{{16}} \times {(16)^{1.5}} = {P_3}$
And hence on doing the simplification,we have
$ \Rightarrow \sqrt {16} \times P = {P_3}$
$ \Rightarrow 4P = {P_3}$
Hence the final pressure is 4P.
Thus the correct answer is option C.

Additional information:
There exist different equations that are better in approximating the behaviour of gas and it is known as the Van der Waals equation that adds two new parameters, i.e. Volume and force of attraction between them, and at lower temperatures, the van der Waals equation reduces to the ideal gas equation.

Note:
At low pressures, no significant force of attraction exists between the gas molecules and they behave like an ideal gas but at higher pressures as the molecules of gas come closer the force of attraction becomes significant, and ideal gas behaviour ceases to exist.