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 :

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$\frac{P}{3}$

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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