Calculate w, q, change in H and change in S for an adiabatic reversible compression of 0.8g of a ideal gas ( molecular weight is 64 g/mol) which occurs at a constant temperature of 25°C from 1.5 atm to 5.2 atm
First, we need to find the initial and final volume of the gas using the ideal gas law:
PV = nRT
Initial conditions:
P1 = 1.5 atm
V1 = ?
n = 0.8g / 64 g/mol = 0.0125 mol
R = 0.0821 L.atm/mol.K
T = 25°C = 298 K
Final conditions:
P2 = 5.2 atm
V2 = ?
n = 0.0125 mol (since the number of moles remains constant)
R = 0.0821 L.atm/mol.K
T = 298 K
Using the ideal gas law, we can find the initial and final volumes:
For initial conditions:
1.5 atm * V1 = 0.0125 mol * 0.0821 L.atm/mol.K * 298 K
V1 = 0.25 L
For final conditions:
5.2 atm * V2 = 0.0125 mol * 0.0821 L.atm/mol.K * 298 K
V2 = 0.68 L
Now that we have the initial and final volumes, we can calculate work done in the adiabatic reversible compression using the formula:
W = -nRT * ln(V2/V1)
W = -0.0125 mol * 0.0821 L.atm/mol.K * 298 K * ln(0.68 L / 0.25 L)
W = -147.61 J
Next, we can calculate the heat transfer using the first law of thermodynamics:
∆U = Q - W
Since the process is adiabatic, Q = 0
Therefore, ∆U = -W
∆U = 147.61 J
Next, we can calculate the change in entropy using the formula:
∆S = nR * ln(V2/V1)
∆S = 0.0125 mol * 0.0821 L.atm/mol.K * ln(0.68 L / 0.25 L)
∆S = 0.043 J/K
Therefore, the work done during the adiabatic reversible compression is -147.61 J, the change in enthalpy is 147.61 J, and the change in entropy is 0.043 J/K.