[bu2012]

One mole of an ideal gas with Cv,m = 3R/2 initially at 325K and 1.50 * 10^5 Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is 2.50 * 10^6 Pa. Calculate the final temperature of the gas. Calculate q, w, delta U, and delta H for this process.

[DevaDevil]

you must show your attempt first, this is forum policy

[bu2012]

Well I know the equations for q, w, delta U, and delta H, but I am unsure of the equation for the first half. I can't find the equation for reversible adiabatic compression, only expansion. I think if I have the equation, I could majority of the problem

[bu2012]

Ok so I found that P_iV_i^Cv,m = P_fV_f^Cv,m. But since volume is held constant, I don't know what to plug in for volume. I know that Pi = 1.50*10⁵ Pa and Pf= 2.50*10⁶ Pa. I can plug these into the equation, but where does the temperature come into play? I'm so lost...please help anyone!

[fledarmus]

How can volume be held constant if you are compressing the gas?

[DevaDevil]

P_iV_i^Cv,m = P_fV_f^Cv,m.

this is only true at constant temperature, which is not the case here.


let's start with the first law of thermodynamics:

dU = dQ - dW

heat loss to surroundings = 0, as it is adiabatic (dQ = 0)
work done at changing volume is: PdV

and

c_v = dU/nT (or: dU = nc_vdT)

this gives us:
 nc_vdT = 0 - PdV

following this you use the ideal gas law (remember, the question stated ideal gas)

pV = nRT; or when differentiated form: VdP + PdV = nRdT
combined:

nc_vdT = - (nRdT - VdP) = VdP - nRdT
use c_v = 3R/2:

3nRdT/2 = VdP - nRdT, or
5nRdT/2 = VdP

use the gas law in its original form to substitute P for V:

5nRdT/2 = (nRT/P)dP = nRT (dP/P)
which gives:
5dT/2T = dP/P

you have start and end pressure, as well as start temperature, so this is solvable for end temperature.

[bu2012]

Thank you! That makes much more sense!