To determine the final pressure, volume, and temperature of the gas, as well as the change in internal energy, energy added to the gas by heat, and work done on the gas during the process, we can follow these steps:
1. Given information:
- Initial pressure (P1) = 100 kPa
- Initial temperature (T1) = 300 K
- Number of moles (n) = 2.00 mol
- Gas constant (R) = 8.314 J/(mol·K)
- Initial volume (V1) is not provided.
2. Using the ideal gas law equation PV = nRT, we can rearrange it to find the initial volume (V1):
- V1 = nRT1 / P1
- V1 = (2.00 mol)(8.314 J/(mol·K))(300 K) / (100 kPa)
- V1 = 49.89 L (rounded to two decimal places)
3. Process 1: The gas is heated at constant pressure to 400 K.
- Final temperature (T2) = 400 K (given)
- Since the pressure remains constant, the final pressure (P2) will also be 100 kPa.
4. To find the final volume (V2) in process 1, we can use the ideal gas law equation again:
- V2 = nRT2 / P2
- V2 = (2.00 mol)(8.314 J/(mol·K))(400 K) / (100 kPa)
- V2 = 66.52 L (rounded to two decimal places)
5. Change in internal energy (ΔU) is given by the equation ΔU = nCvΔT, where Cv is the molar heat capacity at constant volume.
- Since the process is at constant pressure, Cv is not applicable here.
- Therefore, we cannot directly calculate ΔU in this case.
6. Energy added to the gas by heat (q) is given by the equation q = nCpΔT, where Cp is the molar heat capacity at constant pressure.
- Cp for an ideal monatomic gas is 3/2 R.
- Therefore, q = (2.00 mol)(3/2)(8.314 J/(mol·K))(400 K - 300 K)
- q = 249.42 J (rounded to two decimal places)
7. Work done on the gas (w) is given by the equation w = -PΔV.
- Since the pressure remains constant, w = -PΔV = -P(V2 - V1)
- w = -(100 kPa)(66.52 L - 49.89 L)
- w = -1,651 J (rounded to two decimal places)
In summary:
- Final pressure (P2) = 100 kPa
- Final volume (V2) = 66.52 L
- Final temperature (T2) = 400 K
- Change in internal energy (ΔU) = not directly calculable
- Energy added to the gas by heat (q) = 249.42 J
- Work done on the gas (w) = -1,651 J