Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1 . It expands Isothermally …..

Q: Consider one mole of helium gas enclosed in a container at initial pressure P₁ and volume V₁ . It expands Isothermally to volume 4V₁ . After this gas expands adiabatically and its volume becomes 32 V₁ . The work done by the gas during isothermal and adiabatic expansion processes are W_iso and W_adia respectively . If the ratio $\frac{W_{iso}}{W_{adia}} = f ln2 $ , then f is —–

Ans: (1.77 )

Solution: $\frac{P_1}{4}(4V_1)^{5/3} = P_2 (32 V_1)^{5/3}$

$P_2 = \frac{P_1}{4}(\frac{1}{8})^{5/3} = \frac{P_1}{128}$

$\displaystyle W_{adia} = \frac{P_1 V_1 – P_2 V_2}{\gamma -1} $

$\displaystyle W_{adia} = \frac{P_1 V_1 – (P_1/128) (32 V_1)}{\frac{5}{3} -1} $

$\displaystyle W_{adia} = \frac{P_1 V_1 (3/4)}{\frac{2}{3}} $

$\displaystyle W_{adia} = \frac{9}{8}P_1 V_1 $

$\displaystyle W_{iso} = P_1 V_1 ln(\frac{4V_1}{V_1})$

$\displaystyle W_{iso} = 2 P_1 V_1 ln2 $

$\displaystyle \frac{W_{iso}}{W_{adia}} = \frac{16}{9} ln 2 $

f = 16/9 = 1.77