We have a container of a hot ideal monatomic gas. The volume of thecontainer is

Question

We have a container of a hot ideal monatomic gas. The volume of thecontainer is

25 liters. The temperature of the gas is 305 ?C, and itspressure is

0.858 × 10 5 Pa. We allow the gas to cool downto room temperature, which at the time is

21?C. We do not allow the volume of the gas to change. (a) Find thefinal pressure (Pa)

of the gas. (b) Find the amount of heat (J) that passed from thegas to its surroundings

as it cooled. Give a positive value. (c) find the change ofinternal energy (J) of the gas.

Be sure to include the correct sign on the answer.

Explanation / Answer

The number of moles of gas present is needed so we need tofind the volume of gas that would be present at standard conditions P1 V1 / T1 = P2V2 / T2 V2 = (.858 / 1.01) * (273 / (273 + 305)) * 25 =10 liters at standard conditions 10 L / 22.4 L / mole = .446 moles a) final pressure P2 = (T2 /T1) P1 = (294 / 578) * .858 * 10E5 =.436 * 10E5 Pa b) Q = Cv T = .446 * 3 R / 2 *284 = 190 R = 190 * 1.99 = 378 cal c) U = Q + W = Q  since the volume was constant     U = -378 cal   since heat(energy) was removed from the gas V2 = (.858 / 1.01) * (273 / (273 + 305)) * 25 =10 liters at standard conditions 10 L / 22.4 L / mole = .446 moles a) final pressure P2 = (T2 /T1) P1 = (294 / 578) * .858 * 10E5 =.436 * 10E5 Pa b) Q = Cv T = .446 * 3 R / 2 *284 = 190 R = 190 * 1.99 = 378 cal c) U = Q + W = Q  since the volume was constant     U = -378 cal   since heat(energy) was removed from the gas

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