Starting with 3.00 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm

Question

Starting with 3.00 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 23.0 C , a chemist first heats the gas at constant volume, adding 1.52×104J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume.

Calculate the final temperature of the gas.

Calculate the amount of work done by the gas. = 1.35 x 104 J

Calculate the amount of heat added to the gas while it was expanding.

Calculate the change in internal energy of the gas for the whole process.

Explanation / Answer

Specific heat at const pressure =1.04kj/kgK

Specific heat at const volume =0.743Kj/kgk

Acc to first law of themodynamics

Change in heat =work done + change internal energy

nCvdT= 1.52×10^4

3×.743×10^3(T-296)=1.52×10^4

T-296=6.82

T=6.82+296=302.82K

Apply charles law Vi/Ti=Vf/Tf

1/2=302.82/Tf

Tf=605.64K

Amount of work done by gas=n(Cp-Cv)dT=3×0.297×10^3×302.82=2.69×10^5J

Amount of heat added =nCpdT=3×1.04×10^3×302.82=944.8×10^3=9.44×10^5J

Change in internal energy=1.52×10^4+nCvdT=1.52×10^4+3×0.743×10^3×302.82

1.52×10^4+67.5×10^4=69×10^4J

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