Question

According to the first law of thermodynamics, $\Delta U=q+w$. In special cases, the statement can be expressed in different ways. Which of the following is not a correct expression?
[A] At constant temperature, q = -w.
[B] When no work is done, $\Delta U=q$
[C] In gaseous system, $\Delta U=q+P\Delta V$
[D] When work is done by the system, $\Delta U=q+w$

Answer 1

Hint: At constant temperature, the process is isothermal and internal energy is dependent on temperature thus it also becomes zero. For an isochoric process, work done is zero. Heat is released when work is done on the system and is absorbed when work is done by the system.

Complete step by step answer:
In chemistry, we study the law of thermodynamics. In the question it is given to us that $\Delta U=q+w$. We should know that $\Delta U$ is the change in internal energy of the system and it is a function of temperature and volume. ‘q’ is the heat absorbed or released and ‘w’ is the work done.
To answer this question, let us discuss the options one by one.
Firstly, we have that at constant temperature, q = -w. ‘w’ is the work done and ‘q’ is the heat energy. At constant temperature i.e. in an isothermal process as the internal energy is a function of temperature and volume, it becomes zero. Therefore, we can write that,
0 = q + w
or, q = -w.
Therefore, the first statement is correct.

In the second option, we have when no work is done $\Delta U=q$. If ‘w’ becomes zero then the internal energy of the system is equal to the heat. Therefore this statement is also correct.
This process will be an isochoric process because we know work is a function of pressure and volume and it will be zero if there is no change in volume i.e. isochoric process.
In the third option we have, in the gaseous system $\Delta U=q+P\Delta V$.
For a gaseous system, work done is $P\Delta V$ therefore putting this in place of ‘w’ we will get the above equation, this statement is also correct.
And lastly, in the fourth option we have when work is done by the system $\Delta U=q+w$.
When work is done on the system, heat is released i.e. ‘q’ is positive but when work is done by the system, heat is absorbed thus ‘q’ is negative. Therefore, this statement is incorrect.
So, among the given expressions, when work is done by the system, $\Delta U=q+w$ is incorrect.
So, the correct answer is “Option D”.

Note: According to the first law of thermodynamics, if a system is subjected to any cyclic transformation, the sum of heat and work involved is zero.
     \[\oint{dq+\oint{dw=0}}\]
As work done and heat energy both are a path function, their cyclic integration is zero thus, the above statement holds true.