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

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. $$E = f(N,T)$$
• By adding extra molecules, the system will exchange heat to reach the same temperature and the mean thermal energy per a molecule as it had before. The extra energy is the kinetic energy of the extra molecules. – Poutnik Jun 17 at 6:32
• What is your own answer to this question ? An why ? What speaks for and against it ? What speaks for and against each of those 4 answers to be the correct one ? – Poutnik Jun 17 at 10:08
• @Poutnik my assumption that ∆E=f(T) was wrong. Should have been ∆E= f(N,T) if mass is not constant. My other assumption again wrong was, all energy added to system on adding molecules would be transfered to surrounding to keep the process isothermal. I agree with your 1st comment. I'm just asking if my understanding of it is right. And if it is fine, you could probably add an answer. – Vishal prabhu lawande Jun 17 at 10:14
• @Poutnik now my single point against option b,c and d is since mass remaining constant ∆E= f(T). So ∆E= 0 in all those cases. We are only left with option A and that I also agree is right answer. Before reading your 1st comment I had an idea that this question portrays a hypothetical scenario. – Vishal prabhu lawande Jun 17 at 10:25

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.

• Thanks. Could you please help me with "how the addition of molecules doesn't affect the temperature?". Say temperature of system is 400K and we add 1 mole of molecules having 400K, would there be any temperature change in system or any heat transfer? Also if molecules are added at say 500K, would it still be isothermal? – Vishal prabhu lawande Jun 17 at 15:08
• If the addition is at the same pressure and temperature as the original gas, all that happens is that you have more volume and more moles, but the volume per mole stays the same. If you add molecules at 500 K, then heat will be transferred between the old and the new, and the temperature of the original gas is going to rise, while the temperature of added gas is going to drop. – Chet Miller Jun 17 at 15:39

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

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.