Consider the quantity `(MkT)/(pV)` of an ideal gas where `M` is the mass of the gas. It depends on the

Options

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

Answer

To solve the problem regarding the quantity \((MkT)/(pV)\) of an ideal gas, let's break it down step by step.

### Step 1: Understand the Variables
- \(M\) is the mass of the gas.
- \(k\) is the Boltzmann constant.
- \(T\) is the absolute temperature.
- \(p\) is the pressure.
- \(V\) is the volume.

### Step 2: Use the Ideal Gas Law
From the ideal gas law, we know that:
\[ pV = nRT \]
where \(n\) is the number of moles and \(R\) is the universal gas constant.

### Step 3: Relate Molar Mass and Mass
The mass \(M\) of the gas can be expressed in terms of the number of moles \(n\) and the molar mass \(M_0\):
\[ M = n \cdot M_0 \]
where \(M_0\) is the molar mass of the gas.

### Step 4: Substitute \(n\) in Terms of Mass
From the equation \(M = n \cdot M_0\), we can express \(n\) as:
\[ n = \frac{M}{M_0} \]

### Step 5: Substitute \(n\) into the Ideal Gas Law
Substituting \(n\) into the ideal gas law gives:
\[ pV = \left(\frac{M}{M_0}\right)RT \]

### Step 6: Rearranging the Equation
Now we can rearrange this to find \(pV\):
\[ pV = \frac{MRT}{M_0} \]

### Step 7: Substitute into the Original Expression
Now we substitute \(pV\) back into our original expression \(\frac{MkT}{pV}\):
\[ \frac{MkT}{pV} = \frac{MkT}{\frac{MRT}{M_0}} = \frac{MkT \cdot M_0}{MRT} \]

### Step 8: Simplify the Expression
This simplifies to:
\[ \frac{MkT \cdot M_0}{MRT} = \frac{M_0k}{R} \]

### Step 9: Analyze the Result
The final expression \(\frac{M_0k}{R}\) shows that the quantity \(\frac{MkT}{pV}\) depends on the molar mass \(M_0\) of the gas and constants \(k\) and \(R\). Therefore, it is dependent on the nature of the gas.

### Conclusion
The quantity \(\frac{MkT}{pV}\) depends on the molar mass of the gas, which is characteristic of the gas itself.