
For free expansion of a gas in an adiabatic container which of the following is true?
(A) \[\text{Q}\,\text{=}\,W\,=\,0\, and \,\,\Delta U\,=0\]
(B) $\text{Q}\,\text{=}\,\text{0,}\,\text{W}\,\text{}\,\text{0}\,\text{and}\,\text{ }\!\!\Delta\!\!\text{ U}\,\text{=}\,\text{Q}$
(C) $\text{W}\,\text{=}\,\text{0,}\,\text{Q}\,\text{}\,\text{0}\,\text{and}\,\text{ }\!\!\Delta\!\!\text{ U}\,\text{=}\,\text{0}$
(D) $W\,=\,0,\,Q<0\,and\,\Delta U\,=\,0$
Answer
561.6k+ views
Hint: - work done in a closed vessel is zero. In vacuum the external pressure is zero so work done in vacuum zero.
- There is change in temperature in the free expansion of real gas , while ideal gases do not experience the temperature change in the free expansion process.
Complete Solution :
- When in a thermally insulated bottle filled with ideal gas at some temperature and by means of a pipe with a stopper is connected to an insulated bottle which is evacuated. If we suddenly open the stopper, the gas rushes from the first bottle to another bottle until the presser is not equalized. So, this sudden irreversible expanding movement of ideal gas in vacuum is called free expansion. There is no temperature change or heat transfer occurs in this process. The process is adiabatic because of the insulation, so \[\text{ }\!\!\Delta\!\!\text{ Q}\,\text{=}\,\text{0}\] .
- No part of the surrounding moves so the system does not work in the surrounding.
- For ideal gas \[\text{ }\!\!\Delta\!\!\text{ W}\] = work done again external atmosphere, since \[\text{ }\!\!\Delta\!\!\text{ W}\,\text{=}\,\text{P}\Delta \text{V}\] so \[\Delta \text{W}\,\text{=}\,\text{0}\,(P=0)\]
Because, in the vacuum external pressure is zero.
- So from first law of thermodynamics \[\begin{align}
& \Delta U\,=\,Q\,+\,\,\Delta W \\
& \,\Delta U\,\,=\,\,0 \\
\end{align}\]
- The internal energy does not change, which implies that internal energy of an ideal does not depend on the volume at all.
So, the correct answer is “Option A”.
Note: Internal energy of an ideal gas depends only on the temperature.
- If a real gas is suddenly expended by moving the piston outward, there will be a decrease in the temperature of the gas.
- There is change in temperature in the free expansion of real gas , while ideal gases do not experience the temperature change in the free expansion process.
Complete Solution :
- When in a thermally insulated bottle filled with ideal gas at some temperature and by means of a pipe with a stopper is connected to an insulated bottle which is evacuated. If we suddenly open the stopper, the gas rushes from the first bottle to another bottle until the presser is not equalized. So, this sudden irreversible expanding movement of ideal gas in vacuum is called free expansion. There is no temperature change or heat transfer occurs in this process. The process is adiabatic because of the insulation, so \[\text{ }\!\!\Delta\!\!\text{ Q}\,\text{=}\,\text{0}\] .
- No part of the surrounding moves so the system does not work in the surrounding.
- For ideal gas \[\text{ }\!\!\Delta\!\!\text{ W}\] = work done again external atmosphere, since \[\text{ }\!\!\Delta\!\!\text{ W}\,\text{=}\,\text{P}\Delta \text{V}\] so \[\Delta \text{W}\,\text{=}\,\text{0}\,(P=0)\]
Because, in the vacuum external pressure is zero.
- So from first law of thermodynamics \[\begin{align}
& \Delta U\,=\,Q\,+\,\,\Delta W \\
& \,\Delta U\,\,=\,\,0 \\
\end{align}\]
- The internal energy does not change, which implies that internal energy of an ideal does not depend on the volume at all.
So, the correct answer is “Option A”.
Note: Internal energy of an ideal gas depends only on the temperature.
- If a real gas is suddenly expended by moving the piston outward, there will be a decrease in the temperature of the gas.
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