Is it possible to increase the internal energy of an ideal gas in an isothermal process?

Question

Ideal gas is the one in which there are no attractive or repulsive forces acting and hence its internal energy is due to kinetic energy of its molecules. We may then say, for an ideal gas internal energy is a function of temperature##. $$E = f(T)$$ Hence for an ideal gas ∆E=0; for isothermal process.

Thus I thought it is not possible to change internal energy of ideal gas in an isothermal process

Now, I came across the following question.

Q. The internal energy of an ideal gas increases during an isothermal process when gas is
A. Expanded by adding more molecules to it
B. Expanded by adding more heat to it
C. Expanded against zero pressure.
D. Compressed by doing work.

The correct answer given is A. i.e addition of molecules. Definitely adding more molecules would increase the energy of system since internal energy is an extensive property and related as follows $$U = \frac{f}{2}nRT$$ My question is, would system maintain the same temperature even after addition of molecules? My understanding is that, in an isothermal process on addition of energy This energy must be completely used in expansion of gas or must be released into surrounding as heat.$ i.e. $$q = -W $$

If that does not happen, then net energy transfer should lead to change in temperature; i.e. ∆T ≠ 0. making it non-isothermal process.

If my above assumptions are wrong; then; I would be curious to know; where did the energy gained by the system go?

EDIT 1: Summarizing my queries here.

1. Is the above quoted question conveying a hypothetical scenario?
2. If the answer to my main question (In title) is "NO" then question proves to be hypothetical.
3. If question is fine; then how would we justify the increase in internal energy without changing the temperature of system?

After reading some answers and comment section; this is what I feel, I was doing wrong (I've marked what I think are the erroneous statements in my original post in italics and made bold)

1. The following is true only if mass of system is constant. $$E = f(T)$$
2. It would be more appropriate to say; N: number of molecules. $$E = f(N,T)$$

Answer 1

Answer A should be qualified by adding the words "... at constant temperature and pressure." Also, answers B and D can both cause the internal energy of an ideal gas to increase if the process is not carried out isothermally.

Answer 2

First law of thermodynamics is :

I have divided the answer into two parts for better understand of the concept.

For change in internal energy.

$Δ U$ or $\frac{ nfR Δ T}{2}$= $q $ or $c*m*\ Δ T$ ±$ W$ or $PdV$.

Now , for an isothermal process , $Δ T = 0 $ .

Therefore , change in internal energy and heat transfer is always = 0.

But in your answer of the Q , it states that the internal energy can be increased for an isothermal process by adding more molecules to it.

Therefore , For internal energy of a state

$U_2$ - $U_1$ = $q_2$ - $q_1$ ±$ W$ .

If we talk only about single state here. Then , that temperature is not equal = 0 . $T_i$ or $T_f$ may or may not be equal to 0.

To increase the internal energy of a system only. Yes , it can be done by adding more no of molecules. As you can notice , the heat transfer would increase.

I can see how the Q made it confusing to think that if they talk about isothermal process , the reader will mostly think of change in state. Therefore , you can only think of one state where Ti = Tf.

Yes , for each internal energy state. You can increase the internal energy of system.

Commenting upon each option :

A. Expanded by adding more molecules to it
B. Expanded by adding more heat to it
C. Expanded against zero pressure.
D. Compressed by doing work.

A is true.

B is true or similar to A.

C is not true since W = 0.

D is true.

You can make cases here and understood too.

PdV or VdP or $\delta$PV

Answer 3

The answer lies in the fact that when you add more molecules of the gas during the process, the mass, i.e., the number of moles of the gas undergoing the process, changes.

We know, for an ideal gas, its internal energy is a function of its temperature.

$$U=nf/2RT$$

And, change in internal energy is generally represented as,

$$∆U=nf/2R∆T$$

However, in case you add some more molecules of the gas into the system, $n$ also changes. That is,

$$∆U=f/2R∆(nT)$$

Now, coming back to the question of an isothermal process:

In an isothermal process, the temperature always remains constant. Thus, internal energy should also remain constant. Thus, by the First Law of Thermodynamics,

$$∆Q=-∆W$$

And that is fulfilled here, too.

From Thermodynamics, we know that for any process,

$$∆Q=nC∆T$$, where $C$ is the molar heat capacity of the ideal gas for the specific process.

And, $$∆W=n(C-C_v)∆T$$ because, $$∆U=nC_v∆T$$

However, as $n$ changes too in this case, and the process is isothermal, we must write,

$$∆Q=CT∆(n)$$

and, $$∆W=(C-C_v)T∆(n)$$ and, $$∆U=C_vT∆(n)$$

Thus, the process is still isothermal, and $$∆Q=-∆W$$ is still valid, because the internal energy of the gas, in this case, doesn't increase due to supply of heat into the gas, or due to work done on the bas. It increases because the very quantity of the gas undergoing the isothermal process, has changed.