In a thermodynamic process helium gas obeys the law `TP^(-2//5)` = constant,If temperature of `2` moles of the gas is raised from `T` to `3T`, then

Options

A. heat given to the gas is 9RT
B. heat given to the gas is zero
C. increase in internal energy is 6RT
D. work done by the gas is -6RT

Answer

To solve the problem, we need to analyze the thermodynamic process of helium gas obeying the law \( T P^{-2/5} = \text{constant} \). We are given that the temperature of 2 moles of the gas is raised from \( T \) to \( 3T \). We need to find the heat given to the gas, the increase in internal energy, and the work done by the gas.

### Step-by-Step Solution:

1. **Understanding the Process**:
The relationship given is \( T P^{-2/5} = \text{constant} \). This indicates that the process is not isothermal or isobaric, but rather follows a specific path in the \( T-P \) plane.

2. **Using Ideal Gas Law**:
For an ideal gas, we have the equation \( PV = nRT \). For 2 moles of helium, we can express temperature \( T \) as:
\[ T = \frac{PV}{2R} \]

3. **Substituting into the Given Law**:
Substituting \( T \) into the equation \( T P^{-2/5} = \text{constant} \):
\[ \frac{PV}{2R} P^{-2/5} = \text{constant} \]
Simplifying this gives:
\[ P^{1 - 2/5} V = \text{constant} \quad \Rightarrow \quad P^{3/5} V = \text{constant} \]

4. **Identifying the Type of Process**:
The equation \( PV^{5/3} = \text{constant} \) indicates that this is an adiabatic process (since \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \) for a monatomic ideal gas like helium).

5. **Applying the First Law of Thermodynamics**:
In an adiabatic process, the heat exchange \( Q = 0 \). Thus, we have:
\[ \Delta U = Q + W \quad \Rightarrow \quad \Delta U = W \]

6. **Calculating Work Done**:
The work done by the gas can be expressed as:
\[ W = -nC_v \Delta T \]
where \( C_v = \frac{3}{2}R \) for helium. The change in temperature \( \Delta T = 3T - T = 2T \). Therefore:
\[ W = -2 \times \frac{3}{2}R \times 2T = -6RT \]

7. **Calculating Change in Internal Energy**:
The change in internal energy for an ideal gas is given by:
\[ \Delta U = nC_v \Delta T = 2 \times \frac{3}{2}R \times 2T = 6RT \]

8. **Conclusion**:
- Heat given to the gas \( Q = 0 \) (since it is an adiabatic process).
- Increase in internal energy \( \Delta U = 6RT \).
- Work done by the gas \( W = -6RT \).

### Final Answers:
- Heat given to the gas: **0**
- Increase in internal energy: **6RT**
- Work done by the gas: **-6RT**