Question

Consider a reversible isothermal expansion of a gas at temperature τ from volume V to volume V + ∆V . This is not a monatomic ideal gas, but the internal energy of the gas is given by U(τ, V ) = a*V* τ^ 4 , where a is a constant. The pressure is p = (1/3 U)/V . (a) What is the change of energy of the gas in the expansion? (b) How much work is done on the gas in the expansion? (c) How much heat flowed into the gas from the reservoir? (d) What is the pressure after the expansion?

Answer 1

a) change in energy

$\Delta U = Q+W$

As the temperature remains constant in the expansion, $\Delta T =0$ . Therefore, $\Delta U = 0$

b) Work done in a isothermal expansion

$W=nRT \ ln(\frac{final volume}{initial volume})=nRT\ ln(\frac{v+\Delta v}{v})=nRT\ ln(1+\frac{\Delta v}{v})$

c) In an isothermal expansion, the temperature of the gas doesn't change, hence, $\Delta U = 0$

From, first law of thermodynaics, $\Delta U = Q+W$

Q +W = 0

$Q=-W=-nRT\ ln(1+\frac{\Delta v}{v})$

d) Work done in an isothermal expansion, is given by, 2 ways,

$W=nRT \ ln(\frac{final volume}{initial volume})=nRT\ ln(\frac{initial\ pressure}{final\ pressure})$

$\ ln(\frac{final volume}{initial volume})= ln(\frac{initial\ pressure}{final\ pressure})$

$\ ln(1+\frac{\Delta v}{v})= ln(\frac{(1/3U)/v}{Pf})$

substituting for U(τ, V ) = a*V* τ^ 4, we get

$\ ln(1+\frac{\Delta v}{v})= ln(\frac{(1/3)aT^4}{Pf})$

$1+\frac{\Delta v}{v}= \frac{(1/3)aT^4}{Pf}$

FINAL PRESSURE, $Pf= \frac{(1/3)aT^4}{1+\frac{\Delta v}{v}}$