Calculate q, change in U, change in H and change in S for the process of heating 4 mol of O2 adiabatically at a constant pressure of 3.25 atm, from -13°C to 12°C ( Heat capacity of O2 at constant pressure is 37.11 J/K/mol)
Given:
- Initial temperature, T1 = -13°C = 260 K
- Final temperature, T2 = 12°C = 285 K
- Pressure, P = 3.25 atm
- Number of moles, n = 4 mol
- Heat capacity at constant pressure, Cp = 37.11 J/K/mol
To find q (heat transferred), we use the equation:
q = n*Cp*(T2 - T1)
q = 4 mol * 37.11 J/K/mol * (285 K - 260 K)
q = 4 mol * 37.11 J/K/mol * 25 K
q = 3711 J
Therefore, q = 3711 J
To find the change in internal energy (ΔU) for an adiabatic process where q = 0, we can use the first law of thermodynamics:
ΔU = q + W
Since q = 0 for an adiabatic process, ΔU = W
Work done (W) can be calculated using the equation:
W = -PΔV
Since the process is at constant pressure, ΔV = n*R*(T2 - T1) / P
where R is the ideal gas constant (8.314 J/(mol*K))
W = -P * (n*R*(T2 - T1) / P)
W = -4 mol * 8.314 J/(mol*K) * (285 K - 260 K)
W = -4 mol * 8.314 J/(mol*K) * 25 K
W = - 833.6 J
Therefore, ΔU = -833.6 J
Change in Enthalpy (ΔH) can be calculated using the equation:
ΔH = ΔU + PΔV
ΔH = -833.6 J + 3.25 atm * 4 mol * 8.314 J/(mol*K) * (285 K - 260 K) / 1 atm
ΔH = -833.6 J + 3.25 * 4 * 8.314 * 25
ΔH = -833.6 J + 8247.5 J
ΔH = 7413.9 J
Therefore, ΔH = 7413.9 J
Change in Entropy (ΔS) can be calculated using the equation:
ΔS = n*Cp*ln(T2/T1)
ΔS = 4 mol * 37.11 J/K/mol * ln(285 K / 260 K)
ΔS = 4 mol * 37.11 J/K/mol * ln(1.096)
ΔS = 4 mol * 37.11 J/K/mol * 0.092
Therefore, ΔS = 14.4 J/K
In summary:
- q = 3711 J
- ΔU = -833.6 J
- ΔH = 7413.9 J
- ΔS = 14.4 J/K