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06/10/2020 10:39 am
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During isothermal expansion of one mole of an ideal gas from 10 atm to 1atm at 273 K, the work done is [gas constant = 2] :
(a) - 895.8 cal
(b) - 1172.6 cal
(c) - 1257.43 cal
(d) - 1499.6 cal
1 Answer
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06/10/2020 10:42 am
(c) - 1257.43 cal
Explanation:
Work done in expansion of gas
= -2.303 nRT log P1/P2
= -2.303 × 1 × 2 × 273 log 10/1
= - 1257.43 cal