Consider one mole of helium gas enclosed in a container at initial...

Consider one mole of helium gas enclosed in a container at initial pressure

{P}_{1}

and volume

{V}_{1}

. It expands isothermally to volume

4{V}_{1}

. After this, the gas expands adiabatically and its volume becomes

32{V}_{1}

. The work done by the gas during isothermal and adiabatic expansion processes are

{W}_{iso}

and

{W}_{\text{adia}},

\dfrac{{W}_{iso}}{{W}_{adia}} = fln\left(2\right)

, then

f

is _________ . Consider one mole of helium gas enclosed in a container at initial pressure P _ 1 and volume V _ 1 . It expands

Correct answer

Consider one mole of helium gas enclosed in a container at initial pressure P _ 1 and volume V _ 1 . It expands

$∴ \gamma = \dfrac{5}{3}$

In isothermal process,

${W}_{1} = nRTln\left(\dfrac{4V}{V}\right) = 2nRTln\left(2\right)$

In adiabatic process,

$T ( 4V{)}^{\gamma - 1} = {T}^{'} ( 32V{)}^{\gamma - 1}$

$⇒ {T}^{'} = T ( 8{)}^{1 - \gamma} = T \times {8}^{\left(- \dfrac{2}{3}\right)} = \dfrac{T}{4}$

${W}_{2} = - n\dfrac{3}{2}R\left({T}^{'} - T\right) = + \dfrac{3}{2}nR\dfrac{3T}{4} = \dfrac{9}{8}nRT$

$\dfrac{{W}_{1}}{{W}_{2}} = \dfrac{2nRTln2}{\dfrac{9}{8}nRT} = \dfrac{16}{9}\ln 2 = fln2$

$f = \dfrac{16}{9} = 1 . 78$