For the following problem... I found the second pressure by solving P1V1/T1 = P2V2/T2. The Volumes is const so those cancel and then i am able to find the P2. From there i looked at the 1st law and came up with delt U = Qin which = mCvdelta T but that is where I am stuck. Can anyone give me a push?

3.3 A closed rigid tank with a volume of 2 m3 contains hydrogen gas initially at 320 K and 180 kpa. Heat transfer from a reservoir at 500 K takes place until the gas temperature reaches 400 K.
a) Calculate the entropy change (kJ/K) and entropy generation (kJ/K) for the hydrogen gas during the process if the boundary temperature for the gas is the same as the as the gas temperature throughout the process. (0.6264, 0 kJ/K)
b) Determine the entropy generation (kJ/K) for an enlarged system which includes the tank and the reservoir (0.1772 kJ/K)
c) Explain why the entropy generation values differ for parts a) and b)
d) Find the entropy generation for the gas and the total value (kJ/kg) if the system boundary temperature is 450 K throughout the process. (0.1273, 0.1772 kJ/K)
e) The hydrogen gas process is repeated, but paddle-wheel work is used instead of heat transfer. Calculate the entropy generation is this case. (0.6264 kJ/K)
f) Compare the relative irreversibility of processes a), d.) or e)

Here is some help with parts (a) and (b). I hope you will be able to follow the thought process and do most of the rest yourself.

The specific heat of H2 at constant volume is 5/2 R = (5/2)(8.317 J/mol K)/(2.016 mole/g) = 10.31 J/g K = 10.31*10^3 J/ kg K

The mass of H2 present is
M = (2.016 g/mole)*(PV/RT) =
0.273 kg

The entropy added to the gas as a result heat addition is
delta S = Integral of M Cv dT/T
= M Cv ln (T2/T1)
=0.273 kg * 10.31*10^3 J/kg K*ln (400/320)
= 628 K = 0.628 kJ

Since the heat transfer at the boundary of the fluid occurs with no temperature difference at the boundary, the heat transfer process is reversible in this case and there is no net entropy generation for the "universe". That is why the second answer is 0 kJ/K.

In case (b), you are asked what the net entropy generation is if the heat comes from the 500 K reservoir. The heat transferred is
Q = M Cv (delta T = 225.2 kJ
The entropy lost by the reservoir is
Q/Treservoir = 225.2/500 = 0.450 kJ/K

The gas in the tank gains more entropy that the tank loses, and the net entropy gain is 0.628-0.450 = 0.178 kJ/K

To solve this problem, it is important to understand the given information and the relevant equations involved in calculating entropy change and entropy generation.

a) The entropy change for the hydrogen gas can be calculated using the equation:
ΔS = (m * Cv * ln(T2 / T1))

Where:
m is the mass of the hydrogen gas (0.273 kg)
Cv is the specific heat at constant volume for hydrogen gas (10.31 kJ/kg K)
T1 is the initial temperature (320 K)
T2 is the final temperature (400 K)

Plugging in these values into the equation, we can calculate the entropy change for the hydrogen gas:
ΔS = (0.273 kg) * (10.31 kJ/kg K) * ln(400/320)
= 0.628 kJ/K

b) To determine the entropy generation for the enlarged system (including the tank and the reservoir), we need to calculate the heat transferred from the reservoir and compare it with the entropy lost by the reservoir.

The heat transferred can be calculated using the equation:
Q = m * Cv * ΔT

Where:
ΔT is the temperature difference between the reservoir and the gas (ΔT = T_reservoir - T_initial)

Plugging in the values for mass, specific heat, and temperature difference, we can calculate the heat transferred:
Q = (0.273 kg) * (10.31 kJ/kg K) * (500 K - 320 K)
= 225.2 kJ

The entropy lost by the reservoir can be calculated using the equation:
ΔS_reservoir = Q / T_reservoir

Plugging in the values for heat transferred and reservoir temperature, we can calculate the entropy lost by the reservoir:
ΔS_reservoir = 225.2 kJ / 500 K
= 0.450 kJ/K

Finally, to calculate the net entropy generation for the enlarged system, we subtract the entropy lost by the reservoir from the entropy change for the gas:
Entropy generation = ΔS - ΔS_reservoir
= 0.628 kJ/K - 0.450 kJ/K
= 0.178 kJ/K

c) The entropy generation values differ between parts a) and b) because in part a), the heat transfer at the boundary of the fluid occurs with no temperature difference at the boundary, making the process reversible. In contrast, in part b), the heat transfer takes place from a 500 K reservoir to the gas, resulting in a temperature difference and irreversible heat transfer.

d) To find the entropy generation for the gas and the total value, the process boundary temperature should be considered. Using the same method as in part a) and part b), you can calculate the entropy generation for the gas and the total value by substituting the new boundary temperature (450 K) instead of the initial temperature (320 K).

e) For the hydrogen gas process repeated with paddle-wheel work instead of heat transfer, the entropy generation can be calculated similarly to part a) by considering the work done instead of heat transfer. Use the appropriate equation and values to calculate the entropy generation in this case.

f) To compare the relative irreversibility of processes a), d), and e), you can analyze the entropy generation values calculated for each process. A lower entropy generation value indicates a more reversible process with less irreversibility. So, compare the entropy generation values obtained in each case to determine which process has the highest irreversibility.