Question

Which of the following is zero during the adiabatic expansion of the gas?
A. \[{\rm{\Delta T}}\]
B. \[{\rm{\Delta S}}\]
C. \[{\rm{\Delta E}}\]
D. None of these

Answer 1

Hint: A process is said to be adiabatic if there is no heat transfer in a system during any process step. In this process heat, q=0. During the expansion process, the system performs work on the surroundings.

Complete Step by Step Solution:
System is defined as any specified amount of matter under analysis divided from the rest of the universe with a bounding surface.
A system may consist of one or more substances.
In adiabatic expansion, no heat is enabled to come into or take off from the system.
Hence, q=0.
From the first law of thermodynamics, we know that
\[\Delta {\rm{U = q + w}}\]
So, \[\Delta {\rm{U = 0 + w}}\] or \[\Delta {\rm{U = w}}\]


A. \[{\rm{\Delta T}}\]
In an expansion process, work is done by the system on the surroundings.
So, the work done is negative.
Accordingly, internal energy decreases as the work is done at the expense of internal energy.
So, \[{\rm{\Delta E}}\]is negative.
We know that internal energy relies on the movement and configuration of particles which in turn rely on temperature.
So, if \[{\rm{\Delta E}}\] is negative then \[{\rm{\Delta T}}\] is negative.
We see that temperature change is negative and not zero.
So, A is incorrect.

B. \[{\rm{\Delta S}}\]
The entropy change of a system is
\[{\rm{\Delta S = }}\dfrac{{\rm{q}}}{{\rm{T}}}\]
As q=0, \[{\rm{\Delta S}}\]=0.
So, B is correct.

C. \[{\rm{\Delta E}}\]
In this expansion process, work is done by the system on the surroundings.
So, w is negative.
Internal energy decreases as the work is done at the expense of internal energy.
So, \[{\rm{\Delta E}}\] is negative, not zero.
So, C is incorrect.

So, option B is correct.

Note: If there is adiabatic compression, then work is done on the system
by the surroundings. This work is stored in the system in the form of internal energy.
In this case, work done is positive and hence the change in internal energy, \[{\rm{\Delta E}}\] is positive i.e., there would be an increase in internal energy. Due to this, there will be an increase in temperature.