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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.
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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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| V1 | V2 |
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