An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the working substance and operates from a most compressed stage of 100 Nm and
327 K. It expands isothermally to a pressure of 90 Nm and then adiabatically to a
most expanded stage of 27 K. Calculate the AU, 9, and w for each step. Calculate
the net work done and the efficiency of the cycle [Cv, m for the gas) is 25 J/k/mol.​

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

= 100 N·m

= 327 K

= 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln(/) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈ -0.52571 kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln(/) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = -7.5 kJ

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = -5.00574 kJ

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.