Calculate Q, W, ?U, and ?H in each case. Take and

Question

What will be an ideal response?

One mole of an ideal gas, initially at 30°C and 1 bar, is changed to 130°C and 10 bar by three different mechanically reversible processes:

? The gas is first heated at constant volume until its temperature is 130°C; then it is compressed isothermally until its pressure is 10 bar.

? The gas is first heated at constant pressure until its temperature is 130°C; then it is compressed isothermally to 10 bar.

? The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130°C.

Alternatively, take and

Answer 1

Of course, the overall ?U and ?H have to come out the same in every case, since the initial and final states are the same, but the total Q and W should, in general, be different for each case.

a. heating at constant volume followed by compression at constant temperature:

For the constant volume heating, W = 0, so Q = ?U = Cv?T = 5/2*R*(403 K – 303 K) = 2079 J/mol. As usual, ?H = = 7/2*R*(403 K – 303 K) = 2910 J/mol. For the isothermal compression, ?U = ?H = 0, and Q = -W = The initial pressure for this second step is 1 bar * (403 K/303 K) = 1.330 bar.

So, Q = -W = -R*303 K*ln(10/1.330) = -5082.438 J/mol. Adding the two steps gives, for the overall process, Q = 2079 J/mol – 5082.438 J/mol = -3003.438 J/mol, W = 5082.438 J/mol, ?U = 2079 J/mol, and ?H = 2910 J/mol.