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At pressure P and absolute temperature T and 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:

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$\frac{P}{3}$

$P$

$3 P$

$9 P$

answer is C.

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Detailed Solution

If M₀ is molecular mass of the gas then for initial condition PV= $\frac{M}{M_{0}}$ . RT …. (1)
After 2M mass has been added
$P^{'} \frac{V}{3} = \frac{3 M}{M_{0}} . R . \frac{T}{3} ........ (2)$
By dividing (2) by (1)
$P^{'} = 3 P$

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