Answer: sorry for the long answer
To solve this problem, we need to use the isothermal process formula for an ideal gas, which states that the product of pressure and volume is constant at a constant temperature123
We can write this formula as:
P1V1=P2V2
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
We are given the initial pressure P1=80 kPa, the final pressure P2=600 kPa, and the mass of air m=4 kg. We need to find the initial and final volumes, and then use them to calculate the work transfer and heat transfer.
To find the volumes, we can use the ideal gas law, which states that:
PV=mRT
where P is the pressure, V is the volume, m is the mass, R is the specific gas constant, and T is the temperature24
We are given the initial temperature T1=200∘C, which we need to convert to kelvin by adding 273.15. Therefore, T1=473.15 K.
The specific gas constant for air is R=0.287 kJ/kgK4
Plugging these values into the ideal gas law, we can find the initial volume:
V1=P1mRT1
V1=804×0.287×473.15
V1=6.71 m3
Similarly, we can find the final volume by using the same formula with the final pressure and temperature. Since the process is isothermal, the final temperature is equal to the initial temperature, i.e., T2=T1=473.15 K.
Therefore,
V2=P2mRT2
V2=6004×0.287×473.15
V2=0.90 m3
Now that we have both volumes, we can use them to calculate the work transfer during the isothermal process. The work done by a gas in an isothermal process is given by:
W=nRTlnV1V2
where n is the number of moles of gas, and ln is the natural logarithm function23
To find the number of moles of gas, we can use the relation:
n=Mm
where M is the molar mass of gas. For air, the molar mass is M=28.97 g/mol or 0.02897 kg/mol4
Therefore,
n=0.028974
n=138.05 mol
Plugging these values into the work formula, we get:
W=138.05×0.287×473.15ln6.710.90
W=−1149.45 kJ
The negative sign indicates that the work is done by the system on the surroundings, i.e., out of the system.
Finally, we can use the first law of thermodynamics to find the heat transfer during the isothermal process. The first law states that:
ΔU=Q−W
where ΔU is the change in internal energy, Q is the heat transfer, and W is the work transfer14
For an ideal gas undergoing an isothermal process, the change in internal energy is zero, since it depends only on temperature and temperature remains constant. Therefore,
ΔU=0=Q−W
Solving for Q, we get:
Q=W
Substituting the value of W, we get:
Q=−1149.45 kJ
The negative sign indicates that the heat transfer is from the system to the surroundings, i.e., out of the system.
Therefore, the work transfer and heat transfer during this process are both equal to -1149.45 kJ, and both are out of the system.