Question

On heating a gas, pressure and volume both become double. By lowering temperature, one fourth of initial number of moles of air are taken in to maintain the double pressure and volume. Calculate by what fraction, the temperature must have been raised finally.

Answer 1

Initially, `P_(1)=P,V_(1)=V`,
Then after heating, `P_(2)=2P, V_(2)=2V`
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) :. (PV)/(T_(1))=(2Pxx2V)/(T_(2)) " or " T_(2)=4T_(1)`
New P and V are kept constant (double values) when air is taken in. If initialy, npo of moles of air `(n_(1))=n`
Then now no. of moles `=n+(n)/(5)=(5)/(4)n`
Now temperature is raised from 4 `T_(1)` to `T_(f)` (final temperature)
Applying PV=nRT
As P and V are kept constant, `n_(i)T_(i)=n_(f)T_(f)`
`:. " "nxx4T_(1)=(5)/(4)nxxT_(f)" or " T_(f)=(16)/(5)T_(1)`