Please solve Q 76
Dear Student,
Given :
initial volume = 1 dm3 = 1 litre.
final volume = 10 dm3 = 10 litre .
n = 1
R = 8.314 J/mol
T = 273 + 27 = 300 K
as we know that for reversible isothermal process, work done,W = -2.303 nRT log10 V2 / V1
so substituting the given values :
W = -2.303 * 8.314 * 300 *log 10
W = -5744.14 J
Now, if the work done is in joule,then we have to convert it into cal
so, 1 J = 0.000239 K cal
so, W = -5744.14 * 0.000239
= -1.3818 k cal (approximately)
So,the correct answer is (B)
Regards
Given :
initial volume = 1 dm3 = 1 litre.
final volume = 10 dm3 = 10 litre .
n = 1
R = 8.314 J/mol
T = 273 + 27 = 300 K
as we know that for reversible isothermal process, work done,W = -2.303 nRT log10 V2 / V1
so substituting the given values :
W = -2.303 * 8.314 * 300 *log 10
W = -5744.14 J
Now, if the work done is in joule,then we have to convert it into cal
so, 1 J = 0.000239 K cal
so, W = -5744.14 * 0.000239
= -1.3818 k cal (approximately)
So,the correct answer is (B)
Regards