Given quantity of an Ideal gas: MkTpV
For an ideal gas, pV = nRT
Now, we get
MkTpV = MkTnRT
=MknR
we know: n = MMolecularMass
By substituting ′n′ in the above equation:
= (MolecularMass)×MkMR
= MolecularMass×kR
Here, ′R′ and ′k′ are constants, hence the quantity 'MkTpV' depends only on the ′Molecular Mass′ of the gas.
Therefore, 'MkTpV' depends on the nature of the gas.
Hence, the Correct Option is ′D′.