# Rate of change of pressure with volume, adiabatic vs isothermal conditions

1. Aug 13, 2010

### encasements

1. The problem statement, all variables and given/known data
For a particular sample of monatomic gas, the rate of change of pressure with volume under isothermal conditions is -2 x 10^7 Pa m^-3. Calculate the rate of change under adiabatic conditions.

2. Relevant equations
isothermal:
dP/dV = -2 x 10^7
Q = nRTln(Vf/Vi)

monatomic:
PV^(5/3) = constant

dU = dW

3. The attempt at a solution
I feel like this question should be easy but I just don't even know how to approach it. I'd be extreme grateful if anybody could point me in the right direction. Thanks.

2. Aug 13, 2010

### Andrew Mason

1. What is dP/dV in terms of nRT and V for the isothermal process?

2. What is the relationship between P and V in an adiabatic (reversible) process? (hints: it does not involve T and you need to know that it is a monatomic gas).

3. Find dP/dV for the adiabatic process by differentiating the equation in 2.

4. Compare the answers in 1. and 3. using the value for dP/dV that is given for the isothermal process.

AM

3. Aug 13, 2010

### encasements

Thanks for your help. This is what I have so far:

isothermal:
PV = nRT
=> P = (nRT)/V
=> dP/dV = -(nRT)/(V^2)

dP/dV = -2 x 10^7

=> -(nRT)/(V^2) = -2 x 10^7

PV^(5/3) = k
where k is some constant

P = k/(V^(5/3))

=> dP/dV = -(5/3)k/(V^(8/3))

I see that if I had values for n, T, and k I could work out the volume and use that to calculate a value for dP/dV in the adiabatic case -- but since I don't have those values and I'm not sure how to find them I'm afraid I'm still a little stuck.

4. Aug 14, 2010

### Andrew Mason

Well, you know that:

$$k = PV^\gamma$$

and PV = nRT

and

$$nRT/V^2 = 2 \times 10^7$$

AM

5. Aug 17, 2010

### encasements

Thanks for all your help. I've got it now.