# First law of thermodynamics

1. Nov 15, 2007

### El Hombre Invisible

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

A fixed quantity of gas is originally at 20 atmospheres in a volume of 1 litre and a temperature of 300K suddenly expands and comes to equilibrium with a volume of 50 litres and a pressure of 1 atmosphere. If the process is irreversible and involves the transfer of heat and work derive an expression for the difference in the internal energy of the gas in Joules. Take:

$$C_{v} = \frac{5R}{2} mol^{-1}$$

2. Relevant equations

dU = dW + dQ
$$C_{v} = ( \frac{dQ}{dT} )_{V}$$

The specification of Cv (even though the volume isn't constant in this problem) suggests using the state function:

U = U(T,V)

3. The attempt at a solution

Prph! Where to start?!? First I calculated the final value of T using just the equation of state. I got me my partial differential for dU:

$$dU = ( \frac{dU}{dV} )_{T} dV + ( \frac{dU}{dT} )_{V} dT$$

and... that's it. We've only ever been shown how to do reversible processes with one constant so I have no clue where to start with this. Any pointers would be much appreciated.

Cheers,

El Hombre

Last edited: Nov 15, 2007
2. Nov 15, 2007

### Andrew Mason

You need to find the number of moles of gas and the final temperature of the gas, which is simply a matter of applying the ideal gas law.

With that, you can state the change in internal energy of the gas:

$$\Delta U = nC_v\Delta T$$

AM

3. Nov 15, 2007

### El Hombre Invisible

Oops. Accidentally hit the 'report' button instead of 'quote'. That was close. Somebody somewhere would be asking 'How exactly is this offensive...?'

Cheers Andrew for that stroll down memory lane. Haven't seen that equation for a while. It hasn't been brought up again in this or any degree-level course, which makes it rather spurious, but it is question # 3 and the last one we were set was #18, so maybe it's just a filler question to pass the time. I'll go that route because I don't see this as being solveable using TD1, or at least anything we've been taught about it.

Thanks again.

El Hombre

4. Nov 16, 2007

### El Hombre Invisible

Turned out, just in case anyone was following this thread (from my course, for instance) that the aim was to derive that expression using a path in PV space comprising of first a change from the initial isotherm to the final isotherm at constant volume (so no work done) and second a move down that second isotherm to the correct values of P and V. Since the first part of the path is a temperature change only, the 1st law becomes:

$$\Delta U = \left(\frac{dQ}{dT}\right)_{V} \Delta T = C_{V} \Delta T$$

Since the second part is free expansion along an isothermal there is no change in internal energy.

Thanks,

El Hombre

5. Nov 17, 2007

### Andrew Mason

The last sentence is equivalent to: $\Delta U = nC_v\Delta T$ ie. internal energy of a fixed quantity of an ideal gas is a function only of temperature.

You are assuming this to be true and then concluding it is true. If you want to prove that it is true, you have to go to the definition of an ideal gas:

By definition, for an ideal gas:

$$\left(\frac{\partial U}{\partial P}\right)_T = 0$$

Since:

$$\left(\frac{\partial U}{\partial V}\right)_T = \left(\frac{\partial U}{\partial P}\right)_T \left(\frac{\partial P}{\partial V}\right)_T$$

and:

$$\left(\frac{\partial P}{\partial V}\right)_T = -nRT/V^2 \ne 0$$

then:

$$\left(\frac{\partial U}{\partial V}\right)_T = 0$$

Since the first law says:

$$dQ = dU + PdV$$

for a constant volume path:

(1)$$\left(\frac{\partial Q}{\partial T}\right)_V dT = dU + 0$$

The left side is just $nC_vdT$.

Since $$\left(\frac{\partial U}{\partial V}\right)_T = \left(\frac{\partial U}{\partial P}\right)_T = 0$$, (1) must be true for all paths (ie where volume is not constant).

AM

6. Nov 19, 2007

### El Hombre Invisible

Hi Andrew, thanks for the comments.

I'm not sure that was the required answer as we then went on to show how U was independent of P and V for an ideal gas (and not for a real gas of high-ish density). I think the issue here was perhaps my use of the word 'derive' rather than 'state mathematically.' Apologies. However, is your statement of the 1st law correct for an irreversible process like this one?

7. Nov 20, 2007

### Andrew Mason

The first law is a law! It is always true for a thermodynamic system when moving from one equilibrium state to another.

AM

8. Nov 20, 2007

### El Hombre Invisible

I was talking about your statement of it, not the generally true law, in the same way that:

Clearly one cannot say that since a law is a law this is generally true - it is a special case, but is still refered to in this and other sources as a statement of the 1st law for reversible processes. Anyway, what I wanted to check is: is dW = -P dV true for irreversible processes? Judging by Wiki, but in contrast to my lecture notes, the answer is yes, in which case: Thanks for your help, I now have two reasons to wag my finger at our lecturer. ;o)

9. Nov 20, 2007

### Andrew Mason

The first law states that the heat flow in (ie. into or out of) a thermodynamic system is equal to the sum of the work done by (or on) the system and the change in internal energy of the system. That is: dQ = dU + W

A law is always true, not just generally true.

The definition of work is force x displacement: dW = Fds. In a gas, this is equivalent to the external pressure on the gas x change in volume: dW = (F/A)(dsA) = PdV. Since, by convention, we say that dW is positive if work is done by the gas, work is positive if there is an increase in volume. So dW = PdV not -Pdv: dU = dQ - dW = dQ - PdV

The relationship dQ = dU + dW between any two equilibrium states always applies because energy is always conserved.

In a reversible process, the system is in thermodynamic equilbrium at all times. The incremental work done by (or on) a gas is its (internal or external) pressure x change in volume. dW = PdV.

In an irreversible process, the gas pressure may be significantly different than the external pressure. This results in dynamic movement. In such a case, the work done is the external pressure on the gas x change in volume + the kinetic energy produced in the system. [If the internal energy of the system is a combination of thermal and kinetic energy this means it is not in equilibrium]. After the system returns to equilibrium, dQ = dU + PdV will always hold.

AM

10. Nov 22, 2007

### El Hombre Invisible

Same thing. If something is generally true, it is always true.

A ha! Penny's dropped. dW = P dV is NOT true while the system is out of equilibrium (i.e. while the shock wave is propogating) but is true once the system is back to equilibrium. Yeah, that's me not spotting the red-herring in the question: I fixated on the 'irreversible' part and failed to notice the 'comes to equilibrium' part.

Cool, thanks.

El Hombre

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