# One mole of an ideal gas expands from 10 atm against a constant pressure of 1 atm at 27 degree Celsius . The magnitude of work done by the gas is? (R=2 Cal/mol/K)

## $\mathrm{de} g r e e c e l c i u s$

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dk_ch Share
Jan 14, 2017

Given

• $n \to \text{number of moles of the gas} = 1 m o l$
• ${P}_{i} \to \text{Initial pressure of the gas} = 10 a t m$
• ${P}_{f} \to \text{Final pressure of the gas} = 1 a t m$
• $T \to \text{Temperature of the gas} = {27}^{\circ} C = 300 K$

• $R \to \text{Universal gas constant}$
$\text{ "=2calK^-1mol^-1xx4.184J/"cal}$
$\text{ } = 8.368 J {K}^{-} 1 m o {l}^{-} 1$

• $R \to \text{Universal gas constant} = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$

The expansion may be done in two different paths
1) Isothermal reversible
2) Isothemal irrevresible

1) For Isothermal reversible expansion

work done ${w}_{\text{rev}} = - n R T \ln \left({P}_{i} / {P}_{f}\right)$

${w}_{\text{rev}} = - 1 \times 8.368 \times 300 \ln \left(\frac{10}{1}\right) = - 5.78 \times {10}^{3} J$

2) For Isothermal Irreversible expansion

work done ${w}_{\text{irrev}} = - {P}_{f} \times \left({V}_{f} - {V}_{i}\right)$

$\implies {w}_{\text{irrev}} = - {P}_{f} \times \left(\frac{n R T}{P} _ f - \frac{n R T}{P} _ i\right)$

$\implies {w}_{\text{irrev}} = - n R T \left(1 - {P}_{f} / {P}_{i}\right)$

$\implies {w}_{\text{irrev}} = - 1 \times 0.082 \times 300 \left(1 - \frac{1}{10}\right) = - 22.14 L a t m$
$= - 22.14 L a t m \times \frac{101.3 J}{1 L a t m} \approx - 2.242 \times {10}^{3} J$

The second part should be our answer as per given problem.
Since the expansion has been brought about under constant pressure of $1 a t m$. In the first case the pressure has been diminished slowly and reversible from $10 a t m \to 1 a t m$

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