To calculate the heat added to the gas during a reversible process, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
In this case, the gas is monoatomic, which means it has three degrees of freedom and no internal rotational or vibrational energy. Therefore, the change in internal energy (ΔU) can be simplified to ΔU = (3/2) nRΔT, where n is the number of moles and R is the ideal gas constant.
Since the process is carried out at a constant pressure, the work done by the gas is given by:
W = PΔV
where P is the pressure and ΔV is the change in volume.
Now, let's calculate the heat added. We have the following information:
Initial volume (V1) = 0.5 m^3 Final volume (V2) = 1.0 m^3 Pressure (P) = 2 atm Number of moles (n) = 1 mole Ideal gas constant (R) = 8.314 J/(mol·K) (approximately)
ΔV = V2 - V1 = 1.0 m^3 - 0.5 m^3 = 0.5 m^3 ΔT = (ΔU * 2) / (3nR) (from ΔU = Q - W and W = PΔV)
Now, let's calculate ΔT:
ΔT = (ΔU * 2) / (3nR) = ((3/2) nRΔT * 2) / (3nR) = ΔT
Notice that ΔT cancels out on both sides of the equation.
Therefore, the change in temperature (ΔT) is the same as the temperature change of the gas during this process.
Now, to find the heat added (Q), we need to rearrange the first law of thermodynamics equation:
Q = ΔU + W = (3/2) nRΔT + PΔV
Substituting the values we have:
Q = (3/2) * (1 mole) * (8.314 J/(mol·K)) * ΔT + (2 atm) * (0.5 m^3)
Calculating this expression will give you the value of the heat added to the gas during the process.