A heat engine carries 1.70 mol of an ideal monatomic gas around the cycle shown

Question

A heat engine carries 1.70 mol of an ideal monatomic gas around the cycle shown in the figure. Process 1-2 takes place at constant volume and takes the gas from 300 K to 600 K, process 2-3 is adiabatic and takes the gas from 600 K to 455 K, and process 3-1 takes place at a constant pressure and returns the gas to 300 K.

Compute the heat Q for the whole cycle.

Compute the change in internal energy DeltaE for the cycle as a whole.

Calculate the work done by the gas for the cycle.

if the initial pressure at point 1 is 2.2 atm, find the pressure at point 2 in atm.

If the initial pressure at point 1 is 2.2 atm, find the volume at point 3 (in m3).

Explanation / Answer

1 to 2: Q = (3/2) n R delta T = (3/2) * 1.70 * 8.314 * (600-300) = 6360 Joules

2 to 3: Q = 0

3 to 1: Q = (5/2) n R delta T = -5477 J

Total for cycle: Q = 6360 - 5477 = 883 Joules

For the entire cycle, by definition, delta E = 0

The work for the cycle is the same as total Q, 883 J (according to 1st law of thermo)

pressure at 2 = pressure at 1 * temp at 2 / temp at 1 = 2.2*600/300 = 4.4 atm

volume at 3 = nRT / p = 1.70 * 0.08207 * 455 / 2.2 = 28.86 liters = 0.02886 m^3

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