# Homework Help: Simple Thermodynamics

1. Apr 1, 2012

### DirtyLeeds

1. The problem statement, all variables and given/known data

i.) An ideal gas occupies a volume of 2.5 dm3 at a pressure of 0.3 MPa. If the gas is compressed isothermally by a constant external pressure, Pext, so that the final volume is 0.5 dm3, what is the smallest value that the external pressure can have? [2 marks]

ii.)How much work has been done on the gas? [4 marks]

iii.) What is the internal energy change in the gas? [1 mark]

iv.)How much thermal energy has been absorbed by the gas? [2 marks]

v.) What is meant by a reversible process involving an ideal gas? Would the work done on the gas in the above problem be greater if the process had been carried out reversibly? [2 marks]

For part i I just used the equation p1v1=p2v2 and came up with the answer 1.5MPa.

For part ii I used the equation p(Vf-Vi) and got the answer W = -0.3 MJ but i'm not sure if I used the right equation, I originally used the equation W = int pdV but my answer came out as 0. I'm just not sure on the whole on this question, really.

As the process was isothermal, am I right in thinking there was no change in the internal energy of the gas. And because of this the answer for part iv is just 0.3MJ?

Not too sure about part v either. I really need to work on this area...

If anyone could point me in the right direction i'd be thrilled, cheers.

2. Apr 1, 2012

### cepheid

Staff Emeritus
Part i makes sense. The smallest external pressure you could have would be if it were the case that V2 is the final gas volume ie the gas pressure is now equal to the external pressure, and hence the gas will stop compressing.

For part ii, if P is constant, then$$W = \int P\,dV = P \int\,dV = P\Delta V$$so there is no difference between the two methods.

For part iii, yes the internal energy of the gas is constant, because for an ideal gas, internal energy depends only on temperature. Therefore, for part iv, as you stated, all the energy injected into the gas as work must escape as heat.

For part v, you need to hit the books and review reversible processes so that you can answer the question.

3. Apr 1, 2012

### DirtyLeeds

Cheers!

An ideal gas at 0.1 MPa pressure and 298 K expands reversibly and adiabatically from 0.50 dm3 to 1.00 dm3. (For this question, take the molar heat capacity at constant volume to be: CV,m = 12.48 J K-1 mol-1).
Calculate:
(i) its final temperature, [4 marks]

I used the formula Ti/Tf = (Vf/Vi)^x-1 Where x is the ratio of heat capacities, and I assumed that x=1.4, and got the answer 225.84

However, once again I am unsure whether or not i've gone down the right road.

4. Apr 1, 2012

### Andrew Mason

Why are you assuming that the ratio of heat capacities is 1.4? What is Cp? [Hint: You can determine this from the value of Cv - what is Cp-Cv?].

AM