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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?

A

0.003, 275 K

B

0.00043, 261.13 K

C

0.0005, 300 K

D

0.00049, 353 K

Text Solution

Verified by Experts

`w=-2.303 nRT log (v_(2)/v_(1))`
(i) where w is work done by the system under isothermal reversible conditions (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)`…………(i)
Also, `P_(1)V_(1) = P_(2)V_(2)` (at constant temperature)
`therefore 1 xx 10^(7) xx V_(1) = P_(2) xx 10V_(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^(6))`
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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