2 mole of an ideal gas kept at constant temperature 27°C is compressed from 2L to 0.5L .Calculate the work done in the process
Answers
Answered by
1
Answer:
6.9KJ
use formula
At Isothermal Process
work - 2.303nRT (logV2/logv1)
Answered by
2
Answer:
6.9KJ
Explanation:-
Given:- n = 2, T = 27°c + 273 = 300K, Vi = 2L,
Vf = 0.5L.
Work done = 2.303nRT log(Vi/Vf);
= 2.303 × 2 × 8.134 × 300 × log(4);
= 11239 × 0.6020 ;
= 6915
= 6.9KJ.
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