Question

Suppose there is a movable piston insider a cylinder. The piston compresses one mole of a perfect gas, which initially has a pressure at 1 atm insider the cylinder. During the compression, the cylinder is kept in a water bath to maintain the temperature unchanged at 298 K. In this problem the gas constant is 8.3145 J/K mole.

Determine ΔU, ΔH, q and w, when the gas is compressed reversibly to a final pressure of 2 atm.

Answer 1

For an isothermal and reversible compression of a perfect gas from 1 atm to 2 atm at 298 K, the change in internal energy (ΔU) and change in enthalpy (ΔH) are zero, the work done (w) is calculated using the formula w = -nRTln(P2/P1), and the heat exchanged (q) is the negative of the work done.

Step-by-step explanation:

When a perfect gas is compressed reversibly and isothermally from a pressure of 1 atm to 2 atm at 298 K, the change in internal energy (ΔU) is zero because the temperature does not change. According to the first law of thermodynamics, ΔU = q + w, so for an isothermal process involving a perfect gas, q = -w. To compute the work done (w), we can use the formula w = -nRTln(P2/P1), where P1 and P2 are the initial and final pressures, respectively. In this case, w = -(1 mole)(8.3145 J/K mole)(298 K)ln(2 atm / 1 atm). The change in enthalpy (ΔH) for an isothermal process is also zero for an ideal gas because ΔH = ΔU + Δ(pV), and Δ(pV) = nRΔT which is zero for constant temperature. The heat exchanged (q) is simply the negative of the work done, because ΔU is zero