Consider 2 kg of an ideal gas (R=300 J/(kg.K) and Cp=1000 J/(kgK)) contained in a piston cylinder arrangement. The cylinder has a small valve at the end opposite to the piston.The cylinder walls are not adiabatic and allow heat transfer. The initial pressure and volumeare 0.1 MPa and 2 m^3. At time t=0, the gas starts leaking extremely slowly (at theconstant rate of 0.001 kg/s) through the valve (the opening of the valve can be assumed to be controlled to allow a constant leak rate) while the piston is allowed to move. The piston can be considered frictionless. The leakage stops at t=1000s. The whole process can be considered to take place quasistatically with local equilibrium being achieved instantly within the cylinder.The pressure and temperature of the gas do not change during this process.
Answer the following questions.
4A. What is the final volume of the gas (in m3) ?
4B. What is the heat transfer during the process (in joule)?
4C. What is the change in the specific entropy of the gas in the cylinder(in J/(kg.K))?
4D. What is the entropy production rate, in kW/K?
To answer these questions, we need to understand the process and apply the appropriate thermodynamic principles.
Given:
Mass of the gas (m) = 2 kg
Universal gas constant (R) = 300 J/(kg·K)
Specific heat capacity at constant pressure (Cp) = 1000 J/(kg·K)
Initial pressure (P1) = 0.1 MPa
Initial volume (V1) = 2 m^3
Gas leakage rate (dm/dt) = 0.001 kg/s
Time when leakage stops (t2) = 1000 s
4A. What is the final volume of the gas (in m^3)?
To find the final volume (V2) of the gas, we need to consider the mass change during the leakage process. The ideal gas law can be used to relate the initial and final conditions:
P1V1 = P2V2
Rearranging the equation for V2:
V2 = (P1V1)/P2
Given that the pressure of the gas remains constant throughout the process, the final volume can be calculated as:
V2 = (P1V1) / (P2) = (0.1 MPa * 2 m^3) / (0.1 MPa) = 2 m^3
Thus, the final volume of the gas is 2 m^3.
4B. What is the heat transfer during the process (in joules)?
Since the process is quasi-static and the temperature and pressure remain constant, the change in internal energy (∆U) of the gas is zero. Therefore, the heat transfer (Q) during the process can be calculated using the first law of thermodynamics:
Q = ∆U - W
where ∆U is the change in internal energy and W is the work done on the gas.
As ∆U = 0, the heat transfer is simply the work done on the gas. Here, the work done is the work required to overcome the external pressure acting on the gas as it expands due to the leak. The work done is given by:
W = -(P2 - P1) * (V2 - V1)
Substituting the initial and final volumes and pressures into the equation:
W = -(0.1 MPa - 0.1 MPa) * (2 m^3 - 2 m^3) = 0 J
Therefore, the heat transfer during the process is 0 joules.
4C. What is the change in the specific entropy of the gas in the cylinder (in J/(kg·K))?
The specific entropy change (∆s) of the gas can be calculated using the equation:
∆s = Cp * ln(T2/T1) - R * ln(P2/P1)
Since the process is isobaric (constant pressure) and the temperature does not change, ∆s simplifies to:
∆s = R * ln(V2/V1)
Substituting the initial and final volumes into the equation:
∆s = 300 J/(kg·K) * ln(2 m^3 / 2 m^3) = 0 J/(kg·K)
Thus, the change in specific entropy of the gas is 0 J/(kg·K).
4D. What is the entropy production rate, in kW/K?
The entropy production rate (dS/dt) can be calculated using the equation:
dS/dt = dm/dt * [Cp * ln(T2/T1) - R * ln(P2/P1)]
Since the temperature and pressure do not change, the entropy production rate simplifies to:
dS/dt = dm/dt * R * ln(V2/V1)
Substituting the leakage rate and the initial and final volumes:
dS/dt = (0.001 kg/s) * (300 J/(kg·K)) * ln(2 m^3 / 2 m^3) = 0 kW/K
Therefore, the entropy production rate is 0 kW/K.