TPR thermo

[stuw]

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"One mol of helium is equilibrated at STP in a cylinder equipped with a moveable piston of negligible mass. After equilibration the piston is insulated, and a transformation is completed resulting in the temperature of the gas falling to –100˚C, with atmospheric pressure continually maintained. What is the final volume of the gas? Recall that the internal energy of a monatomic ideal gas is given as: E = (3/2)RT. (R = 0.08 L.atm/mol.K)"

Explanation:

"
One mol of helium is equilibrated at STP in a cylinder equipped with a moveable piston of negligible mass. After equilibration the piston is insulated, and a transformation is completed resulting in the temperature of the gas falling to –100˚C, with atmospheric pressure continually maintained. The final volume of the gas is 10.1 L.

Since the question tells us the piston is insulated (q = 0), we know all changes in E are because of pressure-volume work. So ΔE = –PΔV = 3/2 RΔT.

3/2 R(ΔT) = 3/2(0.08)(–100) = (3/2)(–8) = –12

–12 = –PΔV = 1 atm (ΔV)

ΔV = –12 L

And since 1 mol of an ideal gas at STP has 22.4 L of volume, the resultant volume is ~10.4 L, making 10.1 L the best answer. "

My confusion is mostly around this :
–12 = –PΔV = 1 atm (ΔV)
ΔV = –12 L

I understand how they got -12 as the value for change in internal energy, but I don't really understand why the final volume is negative in this case.

 

[brood910]

"One mol of helium is equilibrated at STP in a cylinder equipped with a moveable piston of negligible mass. After equilibration the piston is insulated, and a transformation is completed resulting in the temperature of the gas falling to –100˚C, with atmospheric pressure continually maintained. What is the final volume of the gas? Recall that the internal energy of a monatomic ideal gas is given as: E = (3/2)RT. (R = 0.08 L.atm/mol.K)"

Explanation:

"
One mol of helium is equilibrated at STP in a cylinder equipped with a moveable piston of negligible mass. After equilibration the piston is insulated, and a transformation is completed resulting in the temperature of the gas falling to –100˚C, with atmospheric pressure continually maintained. The final volume of the gas is 10.1 L.

Since the question tells us the piston is insulated (q = 0), we know all changes in E are because of pressure-volume work. So ΔE = –PΔV = 3/2 RΔT.

3/2 R(ΔT) = 3/2(0.08)(–100) = (3/2)(–8) = –12

–12 = –PΔV = 1 atm (ΔV)

ΔV = –12 L

And since 1 mol of an ideal gas at STP has 22.4 L of volume, the resultant volume is ~10.4 L, making 10.1 L the best answer. "

My confusion is mostly around this :
–12 = –PΔV = 1 atm (ΔV)
ΔV = –12 L

I understand how they got -12 as the value for change in internal energy, but I don't really understand why the final volume is negative in this case.

Final volume is ~10.4L not -10.4L.
~ means "around" / "close to".

 

[stuw]

I apologize; scratch "but I don't really understand why the final volume is negative in this case." out. I meant to ask why the change in volume is -12 instead of +12.