A monatomic ideal gas sample is given heat Q. One half of this heat is used as work done by the gas and rest is used for increasing its internal energy. The equation of process in terms of volume and temperature is

Options

A. `(v^(2))/(T^(3))=" constant"`
B. `(v^(2))/(sqrtT)=" constant"`
C. `VT^(3)=" constant"`
D. `V^(2)sqrtT=" constant"`

Answer

To solve the problem, we need to establish a relationship between the volume (V) and temperature (T) of a monatomic ideal gas given that half of the heat supplied (Q) is used for work done (W) and the other half increases the internal energy (ΔU).

### Step-by-step Solution:

1. **Understanding the First Law of Thermodynamics**:
The first law of thermodynamics states that:
\[ Q = W + \Delta U \]
where \(Q\) is the heat supplied, \(W\) is the work done by the gas, and \(\Delta U\) is the change in internal energy.

2. **Given Conditions**:
According to the problem, half of the heat \(Q\) is used for work done and the other half for increasing internal energy:
\[ W = \frac{Q}{2} \quad \text{and} \quad \Delta U = \frac{Q}{2} \]

3. **Expressing Work Done and Change in Internal Energy**:
For a monatomic ideal gas, the change in internal energy can be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \(C_v\) for a monatomic gas is \(\frac{3}{2}R\). Thus:
\[ \Delta U = n \left(\frac{3}{2}R\right) \Delta T \]

The work done by the gas during an expansion can be expressed as:
\[ W = P \Delta V \]

4. **Substituting into the First Law**:
From the first law:
\[ Q = W + \Delta U \implies Q = P \Delta V + n \left(\frac{3}{2}R\right) \Delta T \]
Since \(W = \frac{Q}{2}\) and \(\Delta U = \frac{Q}{2}\), we can write:
\[ \frac{Q}{2} = P \Delta V \quad \text{and} \quad \frac{Q}{2} = n \left(\frac{3}{2}R\right) \Delta T \]

5. **Relating Pressure, Volume, and Temperature**:
For an ideal gas, we have the equation:
\[ PV = nRT \]
Thus, we can express \(P\) as:
\[ P = \frac{nRT}{V} \]

6. **Substituting \(P\) into the Work Done Equation**:
Substitute \(P\) into the work done equation:
\[ W = P \Delta V = \frac{nRT}{V} \Delta V \]
Setting this equal to \(\frac{Q}{2}\):
\[ \frac{nRT}{V} \Delta V = \frac{Q}{2} \]

7. **Using the Change in Internal Energy**:
From the change in internal energy:
\[ n \left(\frac{3}{2}R\right) \Delta T = \frac{Q}{2} \]

8. **Equating the Two Expressions for \(Q\)**:
Since both expressions equal \(\frac{Q}{2}\), we can set them equal to each other:
\[ \frac{nRT}{V} \Delta V = n \left(\frac{3}{2}R\right) \Delta T \]

9. **Separating Variables**:
Rearranging gives:
\[ \frac{T \Delta V}{V} = \frac{3}{2} \Delta T \]
or:
\[ \frac{V}{T} \Delta V = \frac{3}{2} \Delta T \]

10. **Integrating**:
Integrate both sides:
\[ \int \frac{dV}{V} = \frac{3}{2} \int \frac{dT}{T} \]
This gives:
\[ \ln V = \frac{3}{2} \ln T + \ln C \]
where \(C\) is a constant.

11. **Exponentiating**:
Exponentiating both sides results in:
\[ V = C T^{\frac{3}{2}} \]
Rearranging gives:
\[ \frac{V^2}{T^3} = \text{constant} \]

### Final Answer:
The equation of the process in terms of volume and temperature is:
\[ \frac{V^2}{T^3} = \text{constant} \]