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

Question

We have a container of a hot ideal monatomic gas. The volume of the container is 25 liters. The temperature of the gas is 221 C, and its pressure is 0.858 × 105 Pa. We allow the gas to cool down to room temperature, which at the time is 21C. We do not allow the volume of the gas to change. (a) Find the final pressure (Pa) of the gas. (b) Find the amount of heat (J) that passed from the gas to its surroundings as it cooled. Give a positive value. (c) Find the change of internal energy (J) of the gas. Be sure to include the correct sign on the answer

Explanation / Answer

(a) Find the final pressure (Pa) of the gas.

For an ideal gas PV = nRT

Now we have constant volume, hence

P1/P2 = T1/T2

P1 = 0.858e+5 T1 = 273+221 = 4940 K T2 = 273+21 = 2940 K

Final pressure P2 = 0.858x105 x294/494

                              = 0.511x105 Pa

(b) Find the amount of heat (J) that passed from the gas to its surroundings as it cooled. Give a positive value.

Number of moles in the gas

n = PV/RT = 0.858x105 *25x10-3/(8.3*494) {R= 8.3 J/mol-K}

                    = 0.523 mol

For an ideal monoatomic gas Cv = 12.5 J/mol-K

Heat transferred Q = 12.5*0.523*200 = 1307.5 J

(c) Find the change of internal energy (J) of the gas. Be sure to include the correct sign on the answer

Volume is constant and work done is 0

Hence heat transfer is equal to the change internal energy

= -1307.5 J

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