# Internal Energy of Gas and Work done

Tesla In Person
Homework Statement:
Find Internal energy of a Gas
Relevant Equations:
Change in internal energy= Heat added to System - Work done by system
This is a thermodynamics question. A gas absorbs 10 000 J of heat , it releases 3000 J and does 2000 J of work. How much has the internal energy varied?
So I did 10 000 - 3000 -2000 = 5000 J so internal energy decreases by 5000 J. But the correct answer is A) it increased by 5000 J . How ?

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• Delta2

Homework Helper
Gold Member
You have correctly calculated using the first law of thermodynamics that change in internal energy is 5000J. The whole point here is how exactly you interpret the "change in internal energy".
Do you interpret it as "Initial Internal energy minus Final internal energy"?
or "Final internal energy minus Initial internal energy"?

Tesla In Person
You have correctly calculated using the first law of thermodynamics that change in internal energy is 5000J. The whole point here is how exactly you interpret the "change in internal energy".
Do you interpret it as "Initial Internal energy minus Final internal energy"?
or "Final internal energy minus Initial internal energy"?
Change is final - initial so 5000 - 10 000 = -5 000 J. What does the -ve sign mean ?

Homework Helper
Gold Member
Hold on. Change is indeed final-initial but the problem statement doesn't tell you that final is 5000 and initial is 10000. All we can calculate from the problem data is that the change is 5000J , hence change=final-initial=5000J, hence final=5000J+initial, which means that the final internal energy is the initial increased by 5000J.

Last edited:
Tesla In Person
Hold on. Change is indeed final-initial but the problem statement doesn't tell you that final is 5000 and initial is 10000. All we can calculate from the problem data is that the change is 5000J , hence change=final-initial=5000J, hence final=5000J+initial, which means that the final internal energy is the initial increased by 5000J.
Oh i see what the problem was, i assumed 10 000 J was initial energy of the gas. Ok but one thing is still confusing me. How can the internal energy of the gas increase when it looses energy in both transformations?
When it releases 3000 J it looses energy and when it does 2000 J of work it again looses internal energy. So how did the internal energy of the gas increase? How is the final IE higher than initial IE ?

Homework Helper
Gold Member
Oh i see what the problem was, i assumed 10 000 J was initial energy of the gas. Ok but one thing is still confusing me. How can the internal energy of the gas increase when it looses energy in both transformations?
When it releases 3000 J it looses energy and when it does 2000 J of work it again looses internal energy. So how did the internal energy of the gas increase? How is the final IE higher than initial IE ?
Because according to the problem's statement it absorbs 10000J of heat.

• Tesla In Person
Mayhem
Read the question carefully. A way to solve this is by drawing a diagram of how energy is absorbed/released and splitting it into states. Since you have put in effort to answer the question, I'll help you out.

State 0: Initial total energy is some undefined non-zero quantity Q_init.
State 1: Absorbs 10 kJ. In this state, the total energy = Q_init + Q.
State 2: total energy = Q_init + Q - Q_2
State 3: total energy = Q_init + Q - Q_2 - L

This isn't an answer to your question, but it is a way to systematically model what is happening.

Now recall that ##\Delta U = q + w##

Can you solve the rest?

• Tesla In Person
Homework Helper
Gold Member
Understanding the first law is as simple as understanding your bank account balance. After a transaction, the change in your balance follows the simple rule,

change in balance = (what goes in) - (what comes out).

In this case, 10,000 units go in and (2,000 + 3,000) units come out.

• Tesla In Person
Tesla In Person
Thank you all @Delta2 @Mayhem and @kuruman. I understand what i was doing wrong. The 10 kJ was not the initial energy it absorbed that energy and only lost half of that amount (5 kJ) so it still has 5kJ left. So the internal energy increases by 5kJ instead of 10kJ due to the energy lost and work done by gas. 🙏

• Delta2 and kuruman
Tesla In Person
Read the question carefully. A way to solve this is by drawing a diagram of how energy is absorbed/released and splitting it into states. Since you have put in effort to answer the question, I'll help you out.

State 0: Initial total energy is some undefined non-zero quantity Q_init.
State 1: Absorbs 10 kJ. In this state, the total energy = Q_init + Q.
State 2: total energy = Q_init + Q - Q_2
State 3: total energy = Q_init + Q - Q_2 - L

This isn't an answer to your question, but it is a way to systematically model what is happening.

Now recall that ##\Delta U = q + w##

Can you solve the rest?
State 1: Q init + 10 kJ
State 2: Qinit + 10 kJ - 3kJ
State 3 : Qinit + 10 kJ - 3kJ - 2 kJ
Final : Qinit + 5 kJ.
So it increased by 5 kJ.
Thanks !

• Mayhem and Delta2
Mayhem
State 1: Q init + 10 kJ
State 2: Qinit + 10 kJ - 3kJ
State 3 : Qinit + 10 kJ - 3kJ - 2 kJ
Final : Qinit + 5 kJ.
So it increased by 5 kJ.
Thanks !
You got it.