Question

reversible cooling of a perfect gas at constant volume

Answer 1

Answer:

he equation of state for an ideal gas is

pV = muRT

1.

where p is gas pressure, V is volume, mu is the number of moles, R is the universal gas constant (= 8.3144 j/(oK mole)), and T is the absolute temperature. The first law of thermodynamics, the conservation of energy, may be written in differential form as

dq = du + p dV

2.

where dq is a thermal energy input to the gas, du is a change in the internal energy of the gas, and p dV is the work done by the gas in expanding through the change in volume dV.

Constant Volume Process

If V = const., then dV = 0, and, from 2, dq = du; i.e., all the thermal input to the gas goes into internal energy of the gas. We should expect a temperature rise. If the gas has a specific heat at constant volume of CV (j/(oK mole)), then we may set dq = muCV dT. It follows, in this case, that

du = muCV dT

3.

Since du was initially unspecified, we are free to choose its mathematical form. Equation 2 will be retained for du throughout the remainder of the cases.

Constant Pressure Process

If p = const., then dp = 0, and, from 1, p dV = muR dT; i.e., the work done by the gas in expanding through the differential volume dV is directly proportional to the temperature change dT. If the gas has a specific heat at constant pressure of Cp, then dq = muCp dT, and, from 2 (with 3),

muCp dT = muCV dT + muR dT

Simplifying gives an important constitutive relationship between CV, Cp, and R, namely:

Cp = CV + R

4.

Constant Temperature Process

If T = const., then dT = 0, and, from 1, d(pV) = 0, i.e., pressure and volume are inversely proportional. Further, from 2, dq = p dV; i.e., there is no change in internal energy (from 3, du = 0), and all the thermal input to the gas goes into the work of expansion.

Adiabatic Process

If q = const, then dq = 0, and, from 2 (with 3), 0 = muCV dT + p dV; i.e., internal energy of the gas might be reduced in favor of expansion, or vice versa. This expression may be written in an equivalent form as

0 = (CV/R)(dT/T) + dV/V

5.

(division of the first term by muRT, and the second term by pV). Further, from 1,

p dV + V dp = muR dT

or, equivalently,

dp/p + dV/V = dT/T

6.

(division of the Left Hand Side by pV, and the Right Hand Side by muRT).

Equations 5 and 6 may be used to develop relationships between p and V, or p and T:

Explanation: