One mole of an ideal gas is compressed isothermally and reversibly at 607.4 K from 5.60 atm to 8.90 atm.
-Calculate q.

Question

Quizplus · Accepted Answer

The process is isothermal and reversible, so we can use the formula for the work done on or by an ideal gas:

W = -nRT ln(V2/V1)

where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, V1 and V2 are the initial and final volumes of the gas.

Since the process is isothermal, the temperature is constant at 607.4 K. The number of moles is given as one mole. The initial and final volumes are not given, but we can find them using the ideal gas law:

PV = nRT

At the initial pressure of 5.60 atm:

V1 = (nRT)/P1 = (1 mol x 0.08206 L atm K^-1 mol^-1 x 607.4 K)/5.60 atm = 8.89 L

At the final pressure of 8.90 atm:

V2 = (nRT)/P2 = (1 mol x 0.08206 L atm K^-1 mol^-1 x 607.4 K)/8.90 atm = 5.78 L

Substituting these values in the formula for work:

W = -nRT ln(V2/V1)
W = -(1 mol x 0.08206 L atm K^-1 mol^-1 x 607.4 K) ln(5.78 L/8.89 L) 
W = 2.34 kJ

Since the process is reversible, the heat absorbed by the gas (q) is equal to the work done on the gas:

q = -W = -2.34 kJ

Therefore, the correct answer is D) -2.34 kJ.

One Mole of an Ideal Gas Is Compressed Isothermally and Reversibly