Answer:
When a gas undergoes throttling, there is no heat transfer between the gas and its surroundings, and there is no work done by or on the gas. Therefore, the throttling process is considered to be adiabatic and isenthalpic.
In this problem, the initial state of the methane is given as P1 = 1 MPa, T1 = 250 K, and the final pressure is P2 = 100 kPa. Since the process is isenthalpic, we can use the following equation to relate the initial and final temperatures:
H1 = H2
where H is the enthalpy of the gas. Since the kinetic energy of the gas is assumed to be constant, the enthalpy can be approximated as:
H = U + PV
where U is the internal energy of the gas and P and V are its pressure and volume, respectively.
Using the ideal gas law, we can relate the pressure, volume, and temperature of the gas:
PV = nRT
where n is the number of moles of gas and R is the gas constant. Since the mass of the gas is not given, we can assume a convenient value such as 1 kg. The number of moles of methane can then be calculated as:
n = m/M
where m is the mass of the gas and M is its molar mass. For methane, M ≈ 16 g/mol.
Substituting the ideal gas law into the expression for enthalpy, we get:
H = U + nRT
Since the process is isenthalpic, we can set H1 = H2 and simplify:
U1 + nRT1 = U2 + nRT2
The internal energy of an ideal gas is a function of temperature only, so we can write:
U1 = U2
Substituting and rearranging, we get:
T2 = (P2/P1) * T1
T2 = (100 kPa / 1 MPa) * 250 K
T2 ≈ 25 K
Therefore, the exit temperature of the methane after throttling is approximately 25 K.