Final answer:
For an isothermal, isobaric compression of gas, work is calculated by multiplying external pressure by volume change (with proper unit conversion), heat is equal to the work with the opposite sign, internal energy change is zero, and enthalpy change is equal to heat added to the system.
Step-by-step explanation:
To calculate the work (w), heat (q), internal energy change (ΔU), and enthalpy change (ΔH) for an isothermal process under a constant external pressure, we use the first law of thermodynamics and the definition of enthalpy.
The work done on the gas can be calculated using the formula
w = -PextΔV, where Pext is the external pressure and ΔV is the change in volume. Since the external pressure is constant at 100 atm and the volume changes from 20.0 L to 5.00 L, we convert the pressure to the same units as the work (usually joules) by using the conversion factor: 1 atm = 101.325 J/L. The change in volume ΔV is 5.00 L - 20.0 L = -15.00 L (remember that work done on the gas is considered negative when volume decreases).
In an isothermal process for an ideal gas, the internal energy change (ΔU) is zero, as internal energy is a function of temperature, and temperature remains constant. Enthalpy (H) is defined as H = U + PV, so for an isothermal process at constant pressure, the enthalpy change (ΔH) is equal to the heat added to the system since ΔU = 0.
Therefore, ΔH = q and q can be found from ΔH = -w (because ΔU = 0). In other words, the heat exchanged is equal to the work done on the system but with opposite sign, assuming no additional heat is transferred. This is only true for an ideal gas undergoing an isothermal, isobaric process.