1 mole of an ideal gas is compressed isothermally and reversibly at 298 K from a pressure of 1 Pa. To do a work of 100J, to what final pressure should we compress it? (R = 8.314 J/K/mol)

0.96 Pa

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1.0412 Pa

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10 Pa

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96 Pa

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Solution

The correct option is B 1.0412 Pa
$W = - 2.303 n R T l o g \frac{P_{1}}{P_{2}} ∴ l o g \frac{P_{1}}{P_{2}} = \frac{W}{- 2.303 n R T} = \frac{100}{- 2.303 \times 1 \times 8.314 \times 298} = - 0.01753 i . e ., l o g P_{1} - l o g P_{2} = - 0.01753 (T h e u n i t s i n w h i c h P_{1} a n d P_{2} a r e e x p r e s s e d m u s t b e i d e n t i c a l) ∴ - l o g P_{2} = - 0.01753 - l o g P_{1} ∴ - l o g P_{2} = - 0.01753 + l o g P_{1} = 0.01753 + l o g 1 = 0.01753 + 0 l o g P_{2} = 0.01753 \Rightarrow P_{2} = A n t i l o g (0.01753) \Rightarrow P_{2} = 1.0412 P a (S i n c e u n i t o f P_{1} i s P a)$

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