Ideal Gas Expansion with Integration - QS Study

We have one mole of ideal gas at an initial volume of 4.90×103 cm3 which expands to a final volume of 2.45×104 cm3 at a constant temperature T = 298 K. Using your solution from Expanding Gas question “Assuming for the moment the gas is an ideal gas then from the Ideal Gas Law we can obtain the integral below where n is the number of moles of gas in the system and R is the gas constant, R = 8.314 J mol-1 K-1.

 

pV =nRT

p = nRT / V

W = – ʃ [nRT / V] d V

calculate the work done by the gas due to this expansion.

So,

The work done on the gas is calculated below.

w = -nRT ln (Vf / Vi)

w = – (1 mol) (8.314 J K mol-1) (298 K) ln [2.45×104 cm3 /4.90×103 cm3]

w = – 3.99x 103 J = -3.99 k J

The work done on the gas is negative, which means the gas is doing work on the surroundings in the expansion process, the gas is “pushing back’ the surroundings. The question was to calculate the work done by the gas in the expansion which is -w.

– w = + 3.99 kJ