Question

Three samples of the same gas, x, y and z, for which the ratio of specific heats is $ \gamma = \dfrac{3}{2} $ have initially the same volume. The volume of each sample is doubled by adiabatic process in the case of x, by isobaric process in the case of y and by isothermal process in the case of z. If the initial pressure of the sample of x, y and z are in the ratio $ 2\sqrt 2 :1:2 $ then the ratio of their final pressures is
(A) $ 2:1:1 $
(B) $ 1:1:1 $
(C) $ 1:2:1 $
(D) $ 1:1:2 $

Answer 1

Hint: To solve this question we have to obtain the final pressures of all the three samples in terms of the respective initial pressures. This will be done by using the equation of the processes mentioned. Then we will get the required ratio.

Formula Used: The formula used to solve this question is given by,
 $\Rightarrow PV = nRT $ and,
 $\Rightarrow P{V^\gamma } = {\text{Constant}} $
Here $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles, $ T $ is the temperature, and $ R $ is the universal gas constant.

Complete step by step answer
Let the initial pressures of the samples x, y and z be $ {P_x} $ , $ {P_y} $ and $ {P_z} $ respectively. Also, let the respective final pressures be $ {P_x}' $ , $ {P_y}' $ and $ {P_z}' $ .
According to the question we have
 $\Rightarrow {P_x}:{P_y}:{P_z} = 2\sqrt 2 :1:2 $
So, we can write that,
 $\Rightarrow {P_x} = 2\sqrt 2 k $ , …………………..(1)
 $\Rightarrow {P_y} = k $ , and …………………..(2)
 $\Rightarrow {P_z} = 2k $ …………………..(3)
Here $ k $ is a positive constant.
Now, the volume of each sample is doubled. So, for each sample, if $ {V_1} $ and $ {V_2} $ are the initial and the final volumes, then we have
 $\Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = 2 $ …………………..(4)
For sample x:
The process is adiabatic, whose equation is given by
 $\Rightarrow P{V^\gamma } = {\text{Constant}} $
So we have
 $\Rightarrow {P_x}{V_1}^\gamma = {P_x}'{V_2}^\gamma $
Therefore, we get on arranging,
 $\Rightarrow \dfrac{{{P_x}'}}{{{P_x}}} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^\gamma } $
From equations (1) and (4) we can write,
 $\Rightarrow \dfrac{{{P_x}'}}{{2\sqrt 2 k}} = {\left( {\dfrac{1}{2}} \right)^\gamma } $
According to the question $ \gamma = \dfrac{3}{2} $ . Therefore we have on substituting
 $\Rightarrow \dfrac{{{P_x}'}}{{2\sqrt 2 k}} = {\left( {\dfrac{1}{2}} \right)^{3/2}} $
The $ 2\sqrt 2 $ will get cancelled from both sides on the denominator. Hence we have,
 $\Rightarrow {P_x}' = k $ …………………..(5)
For sample y:
The process in this case is isobaric. So we have
 $\Rightarrow {P_y} = {P_y}' $
From (2) we get
 $\Rightarrow {P_y}' = k $ …………………..(6)
For sample z:
The process in this case is isothermal. From the ideal gas equation we have
 $\Rightarrow PV = nRT $
If the temperature $ T $ is constant, then the RHS of the above equation becomes constant. This means that the LHS is also a constant. So for an isothermal process we have
 $\Rightarrow PV = {\text{Constant}} $
This means that
 $\Rightarrow {P_z}{V_1} = {P_z}'{V_2} $
Therefore, we get on arranging,
 $\Rightarrow \dfrac{{{P_z}'}}{{{P_z}}} = \dfrac{{{V_1}}}{{{V_2}}} $
From equations (3) and (4) we get,
 $\Rightarrow \dfrac{{{P_z}'}}{{2k}} = \dfrac{1}{2} $
Hence on calculating we have,
 $\Rightarrow {P_z}' = k $ …………………..(7)
Now, from (5), (6) and (7)
 $\Rightarrow {P_x}':{P_y}':{P_z}' = k:k:k $
Cancelling $ k $ we get
 $\Rightarrow {P_x}':{P_y}':{P_z}' = 1:1:1 $
Thus the ratio of their final pressures is $ 1:1:1 $ .
Hence, the correct answer is option B.

Note
We do not need to get confused by the triple ratio given. We know that at a time we can divide only two numbers. But a triple ratio doesn’t indicate division. It simply indicates that the quantities involved in the triple ratio always are the multiples of the three fixed numbers. So it is always convenient to involve a positive constant to obtain the values of the quantities as multiples of these fixed numbers.