[friedpork]

Hi! I was hoping if somebody can check over my work for a problem set I have due soon. I'm not very fond of thermodynamics so I'm not sure if I am doing it right.

Calculate the heat q when 4.7 litre of an ideal gas at an initial pressure of 48.2 atm is expanded or compressed isothermally to a final volume of 48.2 litres against a constant external pressure of 4.7 atm. Give your answer in J.

w=-pex(V2-V1)
w=-4.7(48.2-4.7) = -204.45 atm L * (8.314 J k-1 mol-1/0.08206 L atm k-1 mol-1) = -20714J
U = q+w since it's isothermal U=0 and therefore q=-w
q= 20714J

Calculate the work w when 5.8 litre of an ideal gas at an initial pressure of 97 atm is expanded or compressed isothermally to a final volume of 97 litres reversibly. Give your answer in J.

w=-nRTln(v2/v1)

PV=nRT
T=PV/nR = (97 * 5. / (1 * 0.08206) = 6855.96K
w=-1*8.314*6855.96*ln(97/5. = -160561.9J

This question seems strange since the temperature is ridiculously high so not entirely sure if I did it correctly.

0.23 litre of an ideal monatomic gas (Cv = 3R/2) initially at 64 °C and 84 atm pressure undergo an expansion against a constant external pressure of 0.93 atm, and do 1.8 kJ of work. The final pressure of the gas is 0.93 atm. Calculate the change in enthalpy, ΔH. Report your answer in J.

ΔH = n* (5/2)R * (T2-T1)

w=-pex(V2-V1)
(w/-pex) + V1 = V2
(-1800/-0.93) * (0.08206/8.314) + 0.23 = V2
V2 = 19.33L

PV=nRT T=PV/nR = (0.93 * 19.33)/ (1 * 0.08206) = 219K

ΔH = 1 * (5/2)*8.314 * (337-219) = -2950 J

Thanks for your time!!

[curiouscat]

Calculate the work w when 5.8 litre of an ideal gas at an initial pressure of 97 atm is expanded or compressed isothermally to a final volume of 97 litres reversibly. Give your answer in J.

w=-nRTln(v2/v1)

PV=nRT
T=PV/nR = (97 * 5. / (1 * 0.08206) = 6855.96K
w=-1*8.314*6855.96*ln(97/5. = -160561.9J

This question seems strange since the temperature is ridiculously high so not entirely sure if I did it correctly.

How'd you know n was 1.

[friedpork]

Ummmm.. because I just assumed if they don't specifically tell you how many moles of a substance there is, you can just put n=1. But I guess that is a really big assumpution.

[curiouscat]

Ummmm.. because I just assumed if they don't specifically tell you how many moles of a substance there is, you can just put n=1. But I guess that is a really big assumpution.

You only need nRT for that equation. Why not just multiply PV?

Assuming n=1 may be causing your bizarre Temps.

[curiouscat]

PV=nRT T=PV/nR = (0.93 * 19.33)/ (1 * 0.08206) = 219K

Again n=1.

This time why not calculate n? You have enough initial state info.

[friedpork]

So I can just go w=-PV ln (v2/v1) = - (97*5.8 ) ln (97/5.8 ) = -1584.8J??? I guess that makes more sense

And then for the second part if I solve for n I got 0.699 mol and then I end up with a delta H of -4213J.