One mole of air, initially at 150°C and 8 bar, undergoes the following mechanically reversible changes. It expands isothermally to a pressure such that when it is cooled at constant volume to 50°C its final pressure is 3 bar. Assuming air is an ideal gas for which and calculate W, Q, ?U, and ?H.

Question

What will be an ideal response?

Answer 1

For the initial state of 150 °C and 8 bar, the molar volume is



If the final state is 50 °C and 3 bar, then the molar volume at the final state is



This is also the molar volume at the intermediate state, since the gas goes from the intermediate state to the final state at constant volume. The temperature at the intermediate state is 150 °C, since the gas goes from the initial state to the

intermediate state isothermally. So, the pressure at the intermediate state is

P = 8.314 Pa

= 392897 Pa = 3.929 bar.

For the isothermal expansion, ?U = ?H = 0 and Q = = -R*423 K *ln(3.929 bar/8 bar) = 2501 J/mol, and

W = -Q = -2501 J/mol.

For the isochoric cooling, ?U = ?T = 2.5 R*(323 K-423 K) = -2079 J/mol, and

?H = ?T = 3.5 R*(323 K – 423 K) = -2910 J/mol. At constant volume, W = 0, so Q = ?U = -2079 J/mol.

For the overall 2-step process, we then have

?U = -2079 J/mol, ?H = -2910 J/mol, Q = 422 J/mol, and W = -2501 J/mol.