[alumina]

Hello everyone,
I've solved this exercise, but I don't know if my reasoning is right.
Can someone point out if I made some mistakes?


A sample of 3.0mol of an ideal gas at 200K and 2.00atm is compressed reversibly and adiabatically until the temperature reaches 250K.
Knowing that the heat capacity at constant volume is 27.5J / (mol * K), calculate Q, W, deltaU, DeltaH, P_f, V_f

Resolution

$$ V_i = \frac{nRT}{P_i} = 24.62L /$$
$$ \Delta U = w = n C_v \Delta T = 4125J /$$
$$ \frac{V_1}{T_1} = \frac{V_2}{T_2}; V_2 = \frac{V_1}{T_1} T_2 /$$
$$ C_p - C_v = R /$$
$$ C_p = R + C_v = 35.8 /$$
$$ \gamma = \frac{C_p}{C_v} = 1.30 /$$
$$ P_1V_1^\gamma = P_2V_2^\gamma /$$
$$ P_2 = 1.49atm /$$
$$ \Delta H = \Delta U + nR\Delta T = 5371.5J /$$
$$ \Delta S = nC_p ln\frac{V_2}{V_1} = 23.94J /$$

[Enthalpy]

If the gas is compressed, its pressure should rise from the initial 2.0atm and its volume decrease.
Apart from that, the reasoning looks sound to me.

If we divide the molar Cv and Cp by R, we get interesting dimensionless values that:
- Are linked with the degrees of freedom of a gaseous molecule (neglecting interactions):
translations, rotations, vibrations, and if hot excitation, splitting, ionization
- Are usable in adiabatic compression:
The temperature ratio, power the first number, gives the volume ratio (inverted if needed)
The temperature ratio, power the second number, gives the pressure ratio

These two numbers differ by 1 and their ratio is gamma (inverted if needed).