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 :

\frac{P}{3}

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P

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3 P

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9 P

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Solution

The correct option is C $3 P$
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)
$P^{'} = 3 P$

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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 :

The correct option is C 3PIf 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

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 :

The correct option is C $3 P$
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)
$P^{'} = 3 P$