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If n moles of an ideal gas are expanded ...

If `n` moles of an ideal gas are expanded is isothermally and reversibly from an initial state in which it has pressure `p_(1)` and volume `V_(1)` to the final state of volume `V_(2)` and pressure `P_(2)`, then

A

`Delta S_(sys) = - 2.303 nR log ((P_(1))/(p_(2)))`

B

`Delta S_(sys) = 2.303 (R )/(n) log ((P_(1))/(p_(2)))`

C

`Delta S_(sys) = - 2.303 (R )/(n) log ((P_(1))/(p_(2)))`

D

`Delta S_(sys) = 2.303 nR log ((P_(1))/(p_(2)))`

Text Solution

Verified by Experts

The correct Answer is:
D
Because the internal energy of an ideal gas is a function of temperature only,
`Delta U = 0`
Therefore, according to the first law of thermodynamics we have
`Delta U = 0 q_("rev") + W`
or `q_("rev") = - W = 2.303 nRT log ((V_(2))/(V_(1)))`
According to Boyle's law,
`p_(1) v_(1) = v_(2) V_(2)`
or `(p_(1))/(p_(2)) = (V_(2))/(V_(1))`
`q_("rev") = 2.303 nRT log ((p_(1))/(p_(2)))`
`Delta S_("sys") = (q_("sys, rev"))/(T)`
`= 2.303 nR log ((p_(1))/(p_(2)))`
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