At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An additional mass 2M of the same gas is added into the container and the volume is then reduced to $\frac{v}{3}$ and the temperature to $\frac{T}{3} .$ The pressure of the gas will now be :

Question

At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An additional mass 2M of the same gas is added into the container and the volume is then reduced to

\frac{v}{3}

and the temperature to

\frac{T}{3} .

The pressure of the gas will now be :

Toppr Admin · Accepted Answer

If M0-xA0-is molecular mass of the gas then for initial condition PV-MM0-RT-1-xA0-After 2M mass has been addedP-x2032-V3-3MM0-R-T3-2-xA0-By dividing -2- and -1-P-x2032-3P

At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An additional mass 2M of the same gas is added into the container and the volume is then reduced to v3 and the temperature to T3. The pressure of the gas will now be :P3P3P9P

If M0 is molecular mass of the gas then for initial condition PV=MM0.RT........(1) After 2M mass has been addedP′.V3=3MM0.R.T3.........(2) By dividing (2) and (1)P′=3P

If $M_{0}$ is molecular mass of the gas then for initial condition PV
$= \frac{M}{M_{0}} . R T . . . . . . . . (1)$
After 2M mass has been added
$P^{'} . \frac{V}{3} = \frac{3 M}{M_{0}} . R . \frac{T}{3} . . . . . . . . . (2)$
By dividing (2) and (1)
$P^{'} = 3 P$