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Q.

If the work done by 2 mol of an ideal gas during isothermal reversible expansion from 5 L to 50 L is -189.1 L atm. At constant pressure, the temperature of the gas (in ° C) is

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a

227

b

127

c

327

d

500

answer is B.

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Detailed Solution

$w = - 2.303 nRTlog [\frac{V_{2}}{V_{1}}] - 189.1 = - 2.303 \times 2 \times 0.0821 \times T \times \log [\frac{50}{5}] - 189.1 = - 2.303 \times 2 \times 0.0821 \times T \times \log 10 T = \frac{189.1}{2.303 \times 2 \times 0.0821} = \frac{189.1}{0.3781} = 500 K or 227 ° C$

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