• Bring your 2026 application questions to our open office hours with Emil Chuck, PhD, Director of Advising Services for HPSA, and get them answered live. Personal statements, secondaries, interview prep, school list strategy. Sunday, May 17 at 9 p.m. Eastern.

TPR thermo

Started by stuw
This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

stuw

Full Member
7+ Year Member
Advertisement - Members don't see this ad
"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.
 
"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".
 
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.