1st Law of Thermo Calculation (Ideal Gas)

[jmm5872]

Hydrogen behaves as an ideal gas at temperatures greater than 200 K and at pressures less than 50 atm.

Suppose 6.00 mol hydrogen is initially contained in a 100-L vessel at a pressure of 2.00 atm. The average molar heat capacity of hydrogen at constant pressure, c_p, is 29.3 J/K/mol in the temperature range of the problem. The gas is cooled reversibly at constant pressure from its initial state to a volume of 50.0-L.

Calculate:
a) Temp of gas in final state
b) Work done on the gas, w, in joules
c) Internal energy change of the gas, $$\Delta$$U, in joules
d) Heat absorbed by the gas, q, in joules

I have worked throught the problem, I am just concerned about the signs of all my answers.

I know that if the volume is decreased and pressure is constant that the temperature will rise. I used PV=nRT to get the initial temp.

I guess my main concern is with the work done. I used this formula:

w = -P_ext$$\Delta$$V

where delta V was 50 - 100 = -50

So I got w = -(2.00 atm)(-50 L) = 100 L atm = 10132.5 J

Then...

q_p = nc_p$$\Delta$$T

$$\Delta$$U = q + w

As I wrote all this down it seems like I have done nothing incorrect.

I think I have been confused about the basic concepts. For some reason I had the notion in my head that if work is positive then q should be negative.

Sorry for the long statement here.

 

[Nino MMP]

I just wanted to explain my problem and why I was confused.a) Temp of gas in final state: The temperature of the gas in the final state will be higher than the initial temperature. Using the ideal gas law, PV=nRT, we can calculate the temperature in the final state as follows:T2 = (P2V2)/(nR) = (2.00 atm)(50.0 L)/(6.00 mol)(8.314 J/K/mol) = 131.3 K b) Work done on the gas, w, in joules:Work is defined as the amount of energy required to move a system from one state to another. In this case, the work done by the gas is equal to the external pressure times the change in volume. Thus, the work done on the gas is given by: w = -Pext\DeltaV = -(2.00 atm)(50.0 L - 100.0 L) = 10132.5 J c) Internal energy change of the gas, \DeltaU, in joules: The internal energy change of the gas is equal to the heat absorbed by the gas plus the work done on the gas. Thus, the internal energy change of the gas is given by: \DeltaU = q + w = (6.00 mol)(29.3 J/K/mol)(131.3 K - 300.0 K) + (-10132.5 J) = -271710.7 J d) Heat absorbed by the gas, q, in joules: The heat absorbed by the gas is equal to the internal energy change of the gas minus the work done on the gas. Thus, the heat absorbed by the gas is given by: q = \DeltaU - w = -271710.7 J - (-10132.5 J) = -261678.2 J