Question

(a) For 2.92 moles of a monatomic ideal gas taken through the cycle in the figure, where V₁ = 5.00V₀ what is W/p₀V₀ as the gas goes from state a to state c along path abc? What is ΔE_int/p₀V₀ in going (b) from b to c and (c) through one full cycle? What is ΔS in going (d) from b to c and (e) through one full cycle?

Answer 1

In the Diagram, the cycle starts at
a- (v0,p0) -----> b- (v1,p0) ----> c- (v1,3p0) ----> a

(a) If W means work per unit mass, then

W = p0 (v1-v0) = 4 p0v0, so w/(p0v0) = 4

(b) The change in internal energy is Cv delta-T,
and delta-T = 5 T0 - T0 = 4 T0 = 4 p0v0/R,
if v0 meant molar volume; and therefore
the change in internal energy is 4 Cv p0v0/R
For a monatomic ideal gas, Cv = 1.5 R, and
the change in internal energy will be 6 p0v0.

(c) If "E(int)" means "internal energy" - though the usual symbol is U -
the change in internal energy through one full cycle is zero,
since internal energy is a state variable.

(d) delta-S going from b to c -
delta-S = integral of dQ/T
The initial temperature (at b) is 5 p0v0/R,
the final temperature (at c) is 15 p0v0/R,
and dQ = Cv dT on this path, as no work is being done.
delta-S = Cv ln|T| to be evaluated at the initial and final T's,
so it's Cv ln(3) = 1.5 R ln(3)

(e) The change in entropy through a full cycle is zero
since entropy is a state variable.