Total kinetic energy of molecules of a gas sample varies `K.E.=(7)/(4)nRT`. This sample contains

Options

A. a. Monatomic gas
B. b. Mixture of amonatomic and diatomic gas
C. c. Mixture of monatomic and polyatomic gas
D. d. Both (2) & (3) are possible

Answer

To solve the problem, we need to analyze the given expression for the total kinetic energy of the gas sample, which is given as:

\[ K.E. = \frac{7}{4} nRT \]

where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

### Step 1: Understand the Kinetic Energy Expression
The kinetic energy of a gas can also be expressed in terms of its degrees of freedom \( F \):

\[ K.E. = \frac{F}{2} nRT \]

### Step 2: Identify Degrees of Freedom for Different Gases
- For a **monoatomic gas**, the degrees of freedom \( F \) is 3 (only translational motion):
\[ K.E. = \frac{3}{2} nRT \]

- For a **diatomic gas**, the degrees of freedom \( F \) is 5 (3 translational + 2 rotational):
\[ K.E. = \frac{5}{2} nRT \]

- For a **polyatomic gas**, the degrees of freedom \( F \) is greater than 5 (includes additional rotational and vibrational modes).

### Step 3: Compare Given Kinetic Energy with Known Forms
The given kinetic energy expression is:

\[ K.E. = \frac{7}{4} nRT \]

We need to determine if this expression can be derived from a combination of monoatomic and diatomic gases or a mixture of monoatomic and polyatomic gases.

### Step 4: Set Up the Equation
Let \( n_1 \) be the number of moles of monoatomic gas and \( n_2 \) be the number of moles of diatomic gas. Then we have:

\[ n_1 + n_2 = n \]

The total kinetic energy can be expressed as:

\[ K.E. = \frac{3}{2} n_1 RT + \frac{5}{2} n_2 RT \]

Substituting \( n_2 = n - n_1 \):

\[ K.E. = \frac{3}{2} n_1 RT + \frac{5}{2} (n - n_1) RT \]

### Step 5: Simplify the Equation
Expanding the equation gives:

\[ K.E. = \frac{3}{2} n_1 RT + \frac{5}{2} n RT - \frac{5}{2} n_1 RT \]

Combining like terms:

\[ K.E. = \left(\frac{3}{2} n_1 - \frac{5}{2} n_1 + \frac{5}{2} n\right) RT \]

This simplifies to:

\[ K.E. = \left(\frac{5}{2} n - 1 n_1\right) RT \]

### Step 6: Set the Kinetic Energy Equal to Given Expression
Now we set this equal to the given kinetic energy:

\[ \frac{5}{2} n - n_1 = \frac{7}{4} n \]

### Step 7: Solve for \( n_1 \)
Rearranging gives:

\[ n_1 = \frac{5}{2} n - \frac{7}{4} n \]

Finding a common denominator (4):

\[ n_1 = \frac{10}{4} n - \frac{7}{4} n = \frac{3}{4} n \]

### Step 8: Determine \( n_2 \)
Now, substituting \( n_1 \) back into the equation for \( n_2 \):

\[ n_2 = n - n_1 = n - \frac{3}{4} n = \frac{1}{4} n \]

### Conclusion
We find that the sample contains:
- \( n_1 = \frac{3}{4} n \) moles of monoatomic gas
- \( n_2 = \frac{1}{4} n \) moles of diatomic gas

Thus, the correct answer is that the sample contains a mixture of monoatomic and diatomic gases, which corresponds to option D (both 2 and 3).