Consider two moles of an ideal diatomic gas at 300 K and 0.507 MPa to ...
Adiabatic Expansion of Ideal Diatomic GasGiven
- Two moles of an ideal diatomic gas
- Initial temperature (T1) = 300 K
- Initial pressure (P1) = 0.507 MPa
- Final pressure (P2) = 0.2033 MPa
- Constant pressure (Pext) = 0.101 MPa
- Final temperature (T2) = ?
- Specific heat capacity at constant volume (cv) = 1.5R
Calculations
Since the process is adiabatic, no heat is exchanged between the system and the surroundings, so Q = 0.
For an ideal gas, the equation of state is given by:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since the process is reversible, we can use:
P1V1 = P2V2
Also, since the process is adiabatic, we have:
P1V1c = P2V2c
where c is the specific heat capacity at constant volume.
Combining the two equations:
P1V1c = P2V2c
P1V1c = P2(P1/P2)c/(c-1)V1c
Simplifying:
V2 = V1(P1/P2)1/c
Substituting into the equation of state:
P1V1 = nRT1
P2V2 = nRT2
we get:
P
1V
1 = P
2V
2(T
2/T
1)