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1mol of an ideal gas undergoes reversibl...

`1mol` of an ideal gas undergoes reversible isothermal expansion form an initial volume `V_(1)` to a final volume `10V_(1)` and does `10kJ` of work. The initial pressure was `1xx 10^(7) Pa`.
a. Calculate `V_(2)`.
b. If there were `2mol` of gas, what must its temperature have been?

Text Solution

Verified by Experts

`w = -2.303n RT log (V_(2)/V_(1))`
(i) where w is work done by the system under isothermal reversible conditionus (work done by the system is negative).
`-10 xx 10^(3)=-2.303 xx 1 xx 8.314 xx T log(P_(1)/P_(2))`
Also, `P_(1)V_(1) = P_(2)V_(2)` (at constant temperature).
`therefore 1 xx 10^(7) xx V_(1) = P_(2) xx 10 V_(1)`
`therefore P_(2) = (1 xx 10^(7))/(10) = 10^(6)` Pa
From Eq.(i)
`-10 xx 10^(3) = -2.303 xx 1 xx 8.314 xx T log 10^(7)/10^(8)`
T = 522.27 K
Using PV = nRT for one mol of gas.
`1 xx 10^(7) xx V_(1) = 1 xx 8.314 xx 522.27`
`therefore V_(1) = 4.34 xx 10^(-4) m^(3)`
(ii) If 2 moles of gas have been used, the temperature would have been `(522.27)/2 = 261.13` K.
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