Calculate $\Delta$S for cooling 2.84 mol of an ideal monatomic gas from 92°C to 42°C at constant volume.

Question

Quizplus · Accepted Answer

We can use the formula: ΔS = nCv ln(T2/T1)

where n is moles of gas, Cv is molar heat capacity at constant volume, T1 and T2 are initial and final temperatures in Kelvin.

Given: n = 2.84 mol, Cv = (3/2)R = 12.5 J/mol*K (for monatomic gas), T1 = 92°C = 365 K, T2 = 42°C = 315 K.

ΔS = 2.84 * 12.5 * ln(315/365) = -5.22 J/K

Now, we can use the formula: ΔS = ΔS_system + ΔS_surroundings = -Q/T + Q/T = 0.

We know that ΔS_surroundings = -ΔH/T, where ΔH is the enthalpy change of the surroundings.

At constant volume, no work is done, so ΔH = Q.

Therefore, -5.22 = -Q/T + Q/T

Q = 5.22 J

Now, we can use the formula: ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat absorbed, and W is the work done.

At constant volume, no work is done, so W = 0.

Therefore, ΔU = Q = 5.22 J

We can use the formula: ΔU = nCvΔT, where ΔT = T2 - T1.

ΔU = 2.84 * 12.5 * (315-365) = -4475 J

Finally, we can use Gibbs free energy equation:

ΔG = ΔH - TΔS

At constant temperature, ΔG = ΔU - TΔS

ΔG = -4475 - 315*(-5.22) = -601.35 J

ΔG = -nRT ln(K)

where K is the equilibrium constant.

ln(K) = -ΔG/(nRT) = 601.35/(2.84*8.314*315) = 0.278

K = e^(0.278) = 1.32

Now, we can use the formula: ΔG° = -RT ln(K)

ΔG° = -8.314*315*ln(1.32) = -2767 J

ΔG° = ΔH° - TΔS°

At constant temperature, ΔG° = ΔU° - TΔS°

ΔU° = ΔH° - PΔV

At constant volume, ΔV = 0.

Therefore, ΔU° = ΔH°

ΔH° = -2767 J

Now, we can use the formula: ΔH = ΣnΔH_f(products) - ΣnΔH_f(reactants)

For an ideal monatomic gas, ΔH_f = 0.

Therefore, ΔH = 0, which means the reaction is adiabatic.

The correct answer is D) -5.22 J/K.

Calculate Δ\DeltaΔ S for Cooling 2