Consider one mole of helium gas enclosed in a container at initial pressure and volume . It expands isothermally to volume . After this, the gas expands adiabatically and its volume becomes . The work done by the gas during isothermal and adiabatic expan

Question

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 $W_{adia}$ , respectively. If the ratio $\frac{W_{iso}}{W_{adia}} = f \ln 2,$ then $f$ is _______. [2020]

Smart Achievers · Accepted Answer

(1.78)

$For monoatomic gas, γ = \frac{5}{3}$

$In adiabatic process$ $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

$\Rightarrow \frac{P_{1}}{4} {(4 V_{1})}^{5 / 3} = P_{2} {(32 V_{1})}^{5 / 3}$

$\Rightarrow P_{2} = \frac{P_{1}}{4} {(\frac{1}{8})}^{5 / 3} = \frac{P_{1}}{128}$

$And W_{adi} = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = \frac{P_{1} V_{1} - \frac{P_{1}}{128} (32 V_{1})}{\frac{5}{3} - 1}$

$= \frac{P_{1} V_{1} (\frac{3}{4})}{\frac{2}{3}} = \frac{9}{8} P_{1} V_{1}$

$In isothermal process,$

$W_{iso} = 2.303 μ R T \log_{10} (\frac{V_{2}}{V_{1}})$

$\Rightarrow W_{iso} = P_{1} V_{1} \ln (\frac{4 V_{1}}{V_{1}}) = 2 P_{1} V_{1} \ln 2$

$∴$ $\frac{W_{iso}}{W_{adia}} = \frac{2 P_{1} V_{1} \ln 2}{\frac{9}{8} P_{1} V_{1}} = \frac{16}{9} \ln 2 = f \ln 2$

$So, f = \frac{16}{9} = 1.7778 \approx 1.78$

Solution

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