Question on thermal. (From Kaplan's online work shop)

[fifi1314]

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Hey guys, this question is from the kaplan online quiz. But the explanation does not consistent with its workshop, any idea? Thanks alot.

One mole ideal gas is trapped in a container covered with a moveable piston. If a 150N force is applied in moving the piston down 5cm, and the compression is isothermal, which of the following is true during this process.
Kaplan's answer:
7.5 J of heat energy leaves the gas. Because it's isothermal, so the T is constant, and so the internal energy doesnt change. 0=Q-W --> Q=W.
Since work is done on the gas, W is negative: W=-Fd=-7.5J
As Q=W, 7.5 J heat energy is lost, it leaves the gas.


But I thought should be 7.5 j heat energy enters the gas. Because if the work is done ON the gas, the gas GAIN energy, and the work is Negative.

 

[hunterpostbacst]

Hey guys, this question is from the kaplan online quiz. But the explanation does not consistent with its workshop, any idea? Thanks alot.

One mole ideal gas is trapped in a container covered with a moveable piston. If a 150N force is applied in moving the piston down 5cm, and the compression is isothermal, which of the following is true during this process.
Kaplan's answer:
7.5 J of heat energy leaves the gas. Because it's isothermal, so the T is constant, and so the internal energy doesnt change. 0=Q-W --> Q=W.
Since work is done on the gas, W is negative: W=-Fd=-7.5J
As Q=W, 7.5 J heat energy is lost, it leaves the gas.


But I thought should be 7.5 j heat energy enters the gas. Because if the work is done ON the gas, the gas GAIN energy, and the work is Negative.

The Kaplan answer is right. If the work is done on the gas, the work should be positive and vice versa. Negative work means that the system lose energy.

 

[sixkiller]

Hey guys, this question is from the kaplan online quiz. But the explanation does not consistent with its workshop, any idea? Thanks alot.

One mole ideal gas is trapped in a container covered with a moveable piston. If a 150N force is applied in moving the piston down 5cm, and the compression is isothermal, which of the following is true during this process.
Kaplan's answer:
7.5 J of heat energy leaves the gas. Because it's isothermal, so the T is constant, and so the internal energy doesnt change. 0=Q-W --> Q=W.
Since work is done on the gas, W is negative: W=-Fd=-7.5J
As Q=W, 7.5 J heat energy is lost, it leaves the gas.


But I thought should be 7.5 j heat energy enters the gas. Because if the work is done ON the gas, the gas GAIN energy, and the work is Negative.

Internal energy for an IDEAL GAS is a function of temperature....

If dT = 0 ; dU = 0

delta U = work + delta q

0 = 7.6 J + delta q

Work done on the system is positive.


The formula for pv work is dw = -pdV....

Therefore, according to the sign convention, work done on the system (compression) is positive and work done by the system (expansion) is negative.

 

[UCB05]

Hey guys, this question is from the kaplan online quiz. But the explanation does not consistent with its workshop, any idea? Thanks alot.

One mole ideal gas is trapped in a container covered with a moveable piston. If a 150N force is applied in moving the piston down 5cm, and the compression is isothermal, which of the following is true during this process.
Kaplan's answer:
7.5 J of heat energy leaves the gas. Because it's isothermal, so the T is constant, and so the internal energy doesnt change. 0=Q-W --> Q=W.
Since work is done on the gas, W is negative: W=-Fd=-7.5J
As Q=W, 7.5 J heat energy is lost, it leaves the gas.


But I thought should be 7.5 j heat energy enters the gas. Because if the work is done ON the gas, the gas GAIN energy, and the work is Negative.

Tricky question... 7.5J of work is done on the system. You're absolutely right that if the gas picks up this potential energy there would be a net increase. HOWEVER, you're told it's an isothermal process so the temperature of the gas doesnt change. Since 7.5J of work is done on the gas, the only way for the process to stay isothermal is for the gas to release the same amount of energy as heat.

 

[fifi1314]

Tricky question... 7.5J of work is done on the system. You're absolutely right that if the gas picks up this potential energy there would be a net increase. HOWEVER, you're told it's an isothermal process so the temperature of the gas doesnt change. Since 7.5J of work is done on the gas, the only way for the process to stay isothermal is for the gas to release the same amount of energy as heat.

Oh I see, your explanation totally make sense, thank you!