# Homework Help: Thermodynamics, the ideal gas equation and temperature change

1. Oct 1, 2011

### Physics_S16

Hi all! I'm new here! I hope my question gets answered because I'm really confused.

1. The problem statement, all variables and given/known data
An amount of 0.18 mol of an ideal gas is held in an insulated cylinder fitted with a piston.

Atmospheric pressure is 1.0 x 10^5 Pa

The volume of the gas is suddenly increased from 1.8 x 10^3 cm3 to 2.1 x 10^3 cm3.

For the expansion of the gas,
(i) calculate the work done by the gas and hence show that the internal energy changes by 30J
(ii)determine the temperature change of the gas and state whether the change is an increase or a decrease

2. Relevant equations
Δu = q + w
pV=nRT?

3. The attempt at a solution
(i) work done by gas= pressure xΔV
So, work done by gas = 30J
Therefore, Δu = -30J

Part (ii) is the problem here

(ii) 3/2 x nRΔT = Δu
So, ΔT = (30x2)/(3x0.18x8.31) = 13.4 K (decrease) ?

However, when using pV=nRT and re-arranging it such that ΔT=(pΔV)/(nR), the
answer is different and it is an increase

Thanks in advance! Any help is much appreciated!

2. Oct 1, 2011

### Andrew Mason

You are using the convention in which W represents the work done ON the gas: dW = -PdV

In this case, the gas does 30J of positive work: $\int PdV = 30 J$. In other words there is NEGATIVE work done ON the gas: $W = -\int PdV = -30 J$.

Since this is adiabatic, the internal energy change must be equal to the work done ON the gas. So $\Delta U = - 30 J$.

You have to apply the first law, determine the change in internal energy and from that determine the change in temperature. You can then use PV=nRT to determine the internal gas pressure. The work is done by the gas against the external pressure. The ideal gas law deals with the internal gas pressure. They are two different things.

AM

3. Oct 2, 2011

### Physics_S16

I am aware that the w represents the work done on the gas and since the gas is expanding (ie. doing work against the atmosphere), w = -pΔV. So Δu = -pΔV + q and since the change is adiabatic, q = 0. Thus, Δu = -pΔV = -30J.

I'm having problem with part(ii) though. How do I find the temperature change?
I can think of two ways.

Method 1
Since the internal energy of an ideal gas is entirely kinetic,
and kinetic energy of n moles of ideal gas = 3/2 x nRΔT
I equate Δu calculated in part(i) with the expression above and then find ΔT.

OR

Method 2
I could use pV = nRT
I calculate T for the two volumes since p, n and R remain constant.
Yet, this way does not make much sense because the higher volume gives a higher T
and the lower volume gives a lower T. By method 1, as the gas expands, the temperature of the gas should fall. Method 2 shows that the temperature of the gas increases.

Why do the two methods give different answers?

4. Oct 2, 2011

### rl.bhat

I could use pV = nRT
Can you use the above expression for the adiabatic change?

5. Oct 2, 2011

### Physics_S16

I don't know. Can I?

6. Oct 2, 2011

### Andrew Mason

You get two different answers because you are measuring two different things.

In the first method the work done by the gas is against the EXTERNAL pressure. This would be the case if the piston was of negligible mass and the initial pressure of the gas was much greater than the external pressure and the final pressure was still greater than the external pressure (ie after the expansion).

In the second method, you are assuming that the change in PV is the external pressure x change in volume. This is not correct. You have to use the internal gas pressure, which is not constant, and you have to also factor in the VdP change:

$$\Delta PV = \int P_{int}dV + \int VdP_{int}$$

AM

7. Oct 2, 2011

### Physics_S16

So, is Method 2 correct for the determination of ΔT?

8. Oct 2, 2011

### Andrew Mason

It is if you know the relationship between P (internal) and V during the process. But you cannot determine that in this case.

In this case, the process is an irreversible adiabatic process, so you cannot use the adiabatic condition. While the process is occurring the system is not in equilibrium. You can't really analyze thermodynamics on a non-equilibrium system. You have to wait until everything settles down (ie. the system reaches equilibrium) and apply equilibrium thermodynamic principles based on the initial and final states. You cannot use PV=nRT because that only applies under equilibrium conditions. P and T are only defined for equilibrium conditions.

AM

9. Oct 2, 2011

### Physics_S16

I'm really sorry... I meant Method 1.

Thank you very much for your time. I really appreciate it!

10. Oct 2, 2011