Question

Consider the quantity $\frac{MkT}{pV}$ of an ideal gas where M is the mass of the gas. It depends on the :

• A. temperature of the gas
• B. volume of the gas
• C. pressure of the gas
• D. nature of the gas

Answer 1

The correct option is D nature of the gas

Given quantity of an Ideal gas: $\frac{MkT}{pV}$

For an ideal gas, $pV$ $=$ $nRT$

Now, we get

$\frac{MkT}{pV}$ $=$ $\frac{MkT}{nRT}$

$=\frac{Mk}{nR}$

we know: $n$ $=$ $\frac{M}{MolecularMass}$

By substituting ${}^{\prime }{n}^{\prime }$ in the above equation:

$=$ $\frac{\left(MolecularMass\right)\times Mk}{MR}$

$=$ $\frac{MolecularMass\times k}{R}$

Here, ${}^{\prime }{R}^{\prime }$ and ${}^{\prime }{k}^{\prime }$ are constants, hence the quantity '$\frac{MkT}{pV}$' depends only on the ${}^{\prime }Molecular$ $Mas{s}^{\prime }$ of the gas.

Therefore, '$\frac{MkT}{pV}$' depends on the nature of the gas.

Hence, the Correct Option is ${}^{\prime }{D}^{\prime }$.