An ideal has ( Cp,m= 28.8 J/mol/k) is expanded reversibly and adiabatically from a volume of 2.0 dm-3 at a pressure of 4 ATM and temperature of 298k, to a volume of 4 dm-3. Calculate;
a). Final temperature and pressure of the gas.
b). q, w, change in U and change in H for the process
To solve this problem, we can use the equations for an adiabatic process:
a) Final temperature:
From the ideal gas law, we have:
P1V1/T1 = P2V2/T2
Given:
P1 = 4 ATM
V1 = 2.0 dm^3
T1 = 298 K
V2 = 4 dm^3
Rearranging the equation:
T2 = (P2V2T1) / (P1V1)
Plugging in the values:
T2 = (4 * 4 * 298) / (2 * 4)
T2 = 298 K
Final temperature of the gas is 298 K
Final pressure:
Using the equation for an adiabatic process:
P1V1^gamma = P2V2^gamma
Where gamma = Cp/Cv = 28.8 J/mol/K / (28.8 J/mol/K - 8.31 J/mol/K) = 1.41
Rearranging the equation:
P2 = P1 * (V1/V2)^gamma
Plugging in the values:
P2 = 4 * (2/4)^1.41
P2 = 2.83 ATM
Final pressure of the gas is 2.83 ATM
b) Calculate q, w, change in U, and change in H:
Since the process is adiabatic, q = 0
For an adiabatic process, w = - change in U
w = -Cp * (T2 - T1) = -28.8 * (298 - 298) = 0
Change in U:
ΔU = 0 because w = 0
Change in H:
ΔH = ΔU + Δng * (T2 - T1)
Since the process is isothermal, ΔT = 0
ΔH = 0
Therefore:
a) Final temperature of the gas is 298 K
Final pressure of the gas is 2.83 ATM
b) q = 0
w = 0
Change in U = 0
Change in H = 0