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Question
Consider the quantity \[\frac{MkT}{pV}\] of an ideal gas where M is the mass of the gas. It depends on the
Options
temperature of the gas
volume of the gas
pressure of the gas
nature of the gas.
Solution
nature of the gas.
\[\text { In an ideal gas, the equation of state is given by }\] \[PV = nRT\] \[ \Rightarrow PV = n N_A \frac{R}{N_A}T\]
\[ \Rightarrow PV = n N_A kT\]
\[ \Rightarrow \frac{1}{n N_A} = \frac{kT}{PV}\]
\[\text { Multiplying both sides by mass of the gas M, we get }\] \[\frac{M}{n N_A} = \frac{MkT}{PV}\] \[\text { Now, n N_A gives the total number of molecules of the gas . }\] \[\text { Also }, \frac{M}{n N_A} \text{ gives the mass of a single molecule }. \] \[\text { Hence, }\frac{MkT}{PV} \text { is the mass of a single molecule of the gas, } \] \[\text { Molecular mass is a property of the gas .} \]
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