One mole of an Ideal gas was compressed isothermally and reversibly from 22-4 - 10-3 m3 and one atmospheric pressure to 11-2 - 10-3 m3 R-8-314 J-K-mol- The work done on the system is

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Standard XIII

Chemistry

Work Done in Isothermal Reversible Process

One mole of a...

Question

One mole of an Ideal gas was compressed isothermally and reversibly from $22.4 \times 10^{- 3} m^{3}$ and one atmospheric pressure to $11.2 \times 10^{- 3} m^{3} (R = 8.314 J / K / m o l)$ . The work done on the system is

- 1134 J

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1134 J

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1573 J

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Solution

The correct option is C 1573 J
Since 1 mole of an Ideal gas at 1 atmosphere was taken and since initial volume was $22.4 d m^{3}$ , the temperature must be 273 K. (Molar volume of an ideal gas at STP is $22.4 d m^{3}$ ) $W = - 2.303 n R T l o g (\frac{V_{2}}{V_{1}}) = - 2.303 n R T l o g (\frac{11.2 \times 10^{- 3}}{22.4 \times 10^{- 3}}) = - 2.303 n R T l o g (0.5) = - 2.303 \times 1 m o l \times 8.314 J / K / m o l \times 273 K \times (- 0.3010) W = + 1573 J$

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One mole of an Ideal gas was compressed isothermally and reversibly from 22.4×10−3m3 and one atmospheric pressure to 11.2×10−3m3(R=8.314J/K/mol). The work done on the system is

One mole of an Ideal gas was compressed isothermally and reversibly from $22.4 \times 10^{- 3} m^{3}$ and one atmospheric pressure to $11.2 \times 10^{- 3} m^{3} (R = 8.314 J / K / m o l)$ . The work done on the system is