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Question: A gas for which \(\gamma =1.5\) is suddenly compressed to \(\dfrac{1}{4}\)th of its initial value th...

A gas for which γ=1.5\gamma =1.5 is suddenly compressed to 14\dfrac{1}{4}th of its initial value then the ratio of the final to the initial pressure is
A. 1:16
B. 1:8
C. 1:4
D. 8:1

Explanation

Solution

Since the compression happening in the given process is sudden, we could take it as the clear indication of adiabatic process. Now we can recall the relation of initial and final pressures and temperatures under adiabatic process. We are given that the final volume is 14\dfrac{1}{4} th of the initial volume and also the value of γ, substituting these values in the known equation will give you the required equation.

Formula used:
For an adiabatic process, PVγP{{V}^{\gamma }} is constant.
P1V1γ=P2V2γ{{P}_{1}}{{V}_{1}}^{\gamma }={{P}_{2}}{{V}_{2}}^{\gamma }

Complete answer:
We are given a gas which is suddenly compressed to 14\dfrac{1}{4} th of its initial volume. We are asked to find the ratio of the final pressure to the initial pressure. Since we are given that the gas is suddenly compressed, we could think of a process for which sudden compression takes place. Without doubt, we can say that the process is adiabatic. Let us recall what happens in an adiabatic process.
In an adiabatic process, the energy transfer from the thermodynamic system to surroundings is only in the form of work, that is, there is no transfer of heat and matter in an adiabatic process. Also, an adiabatic process, irrespective of whether it is an expansion or compression is sudden, so that there is no time for the exchange of heat as a resultant of which there will always be a change in temperature.
We are also given γ which the ratio of specific heats at constant pressure and at constant volume as,
γ=1.5\gamma =1.5 ………………………. (1)
Let the gas undergo adiabatic change from (P1,V1)\left( {{P}_{1}},{{V}_{1}} \right) to(P2,V2)\left( {{P}_{2}},{{V}_{2}} \right), we are given that,
V2=V14{{V}_{2}}=\dfrac{{{V}_{1}}}{4} ………………………. (2)
Also, for an adiabatic process we have PVγP{{V}^{\gamma }} as a constant. Therefore,
P1V1γ=P2V2γ{{P}_{1}}{{V}_{1}}^{\gamma }={{P}_{2}}{{V}_{2}}^{\gamma } ………………………………… (3)
Now we can substitute equations (1) and (2) in equation (3).
P1V11.5=P2(V14)1.5{{P}_{1}}{{V}_{1}}^{1.5}={{P}_{2}}{{\left( \dfrac{{{V}_{1}}}{4} \right)}^{1.5}} ……………… (4)
Rearranging (4) we get,
P2P1=((V14)V1)1.5\dfrac{{{P}_{2}}}{{{P}_{1}}}={{\left( \dfrac{\left( \dfrac{{{V}_{1}}}{4} \right)}{{{V}_{1}}} \right)}^{1.5}}
P2P1=(4)1.5=(2)3=8\dfrac{{{P}_{2}}}{{{P}_{1}}}={{\left( 4 \right)}^{1.5}}={{\left( 2 \right)}^{3}}=8
P2=8P1{{P}_{2}}=8{{P}_{1}}
Therefore, the ratio of the final to the initial pressure is 8:1.

Hence, the answer to the question is option D.

Note: For solving questions involving thermodynamic processes, firstly, we have to identify which process is discussed in the question. Sometimes, if you are lucky, the process will be directly mentioned in the question in other cases you have to identify them from the keywords given. Though the question doesn’t directly mention that adiabatic process is discussed here, it does say that the compression is sudden. So the keyword here is ‘suddenly’.