A box contains n molecules of a gas. How will | vedclass

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(C) We know that from the ideal gas equation,$PV = n_{moles}RT$.Here,$n_{moles} = \frac{N}{N_{A}}$,where $N$ is the number of molecules of the gas and

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Pressure will be doubled

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(C) We know that from the ideal gas equation,$PV = n_{moles}RT$.Here,$n_{moles} = \frac{N}{N_{A}}$,where $N$ is the number of molecules of the gas and $N_{A}$ is the Avogadro's number.Substituting the value of $n_{moles}$ into the ideal gas equation,we get $PV = \frac{N}{N_{A}}RT$.This implies that the pressure $P$ of the gas is directly proportional to the number of molecules $N$ of the gas,provided the temperature $T$ and volume $V$ remain constant $(P \propto N)$.Therefore,if the number of molecules is doubled $(N' = 2N)$,the pressure will also be doubled $(P' = 2P)$.

$A$ box contains $n$ molecules of a gas. How will the pressure of the gas be affected if the number of molecules is made $2n$?

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