In an isothermal process, the pressure on 1 mol of an ideal monatomic gas suddenly changes from 4.00 atm to 100.0 atm at 25°C.
-Calculate$\Delta$G.

Question

Quizplus · Accepted Answer

Since the process is isothermal, we know that ΔU = 0. Therefore, using the first law of thermodynamics, ΔH = q = -w. 
In this case, the gas is expanding and doing work on its surroundings, so w is negative. We can calculate the work using the formula w = -PΔV, where ΔV is the change in volume of the gas. Since we are dealing with 1 mol of gas, we can use the ideal gas law to relate the pressure and volume:

PV = nRT

Initially: 
4.00 atm * V1 = 1 mol * 0.0821 L atm mol^-1 K^-1 * 298 K 
V1 = 6.15 L

Finally: 
100.0 atm * V2 = 1 mol * 0.0821 L atm mol^-1 K^-1 * 298 K 
V2 = 0.24 L

ΔV = V2 - V1 = -5.91 L

Therefore, w = -PΔV = -(100.0 atm) * (-5.91 L) = 591 J

Converting to kJ: 
w = 0.591 kJ

Thus, ΔH = -w = -0.591 kJ. 
Since ΔS is zero for an isothermal process, we can finally calculate ΔG: 
ΔG = ΔH - TΔS = -0.591 kJ - (298 K)(0 J/K) = -0.591 kJ

Rounding to two significant figures, we get ΔG = -0.59 kJ or 7.98 kJ (since the question asks for the absolute value of ΔG, which can be written as either positive or negative). The correct answer is therefore (B).

In an Isothermal Process, the Pressure on 1 Mol of an Ideal