An ideal monatomic gas undergoes a process where its pressure is inversely proportional to its temperature.
(a) Calculate the molar specific heat of the process.
(b) Find the work done by two moles of gas if the temperature change from `T_1` to `T_2`.

Explanation

To solve the problem step by step, we will address both parts of the question regarding the ideal monatomic gas.

### Part (a): Calculate the molar specific heat of the process.

1. **Understand the relationship between pressure and temperature**:
Given that pressure \( P \) is inversely proportional to temperature \( T \), we can express this as:
\[ P \propto \frac{1}{T} \implies P \cdot T = \text{constant} \]

2. **Use the ideal gas equation**:
The ideal gas equation is given by:
\[ PV = nRT \]
Rearranging this gives:
\[ T = \frac{PV}{nR} \]

3. **Combine the relationships**:
From the relationship \( P \cdot T = \text{constant} \) and the ideal gas equation, we can derive that:
\[ P^2 V = \text{constant} \]

4. **Identify the type of process**:
This indicates a polytropic process of the form:
\[ PV^x = \text{constant} \]
Here, we can identify \( x = \frac{1}{2} \).

5. **Calculate the molar specific heat**:
The formula for molar specific heat in a polytropic process is:
\[ C = C_v + \frac{R}{1 - x} \]
For a monatomic gas, \( C_v = \frac{3R}{2} \). Substituting \( x = \frac{1}{2} \):
\[ C = \frac{3R}{2} + \frac{R}{1 - \frac{1}{2}} = \frac{3R}{2} + 2R = \frac{3R}{2} + \frac{4R}{2} = \frac{7R}{2} \]

### Answer for Part (a):
The molar specific heat of the process is:
\[ C = \frac{7R}{2} \]

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### Part (b): Find the work done by two moles of gas if the temperature changes from \( T_1 \) to \( T_2 \).

1. **Use the first law of thermodynamics**:
The first law states:
\[ \Delta U = Q - W \]
Rearranging gives:
\[ W = Q - \Delta U \]

2. **Express heat \( Q \)**:
The heat added to the system can be expressed as:
\[ Q = nC\Delta T \]
where \( \Delta T = T_2 - T_1 \).

3. **Express change in internal energy \( \Delta U \)**:
The change in internal energy for an ideal gas is given by:
\[ \Delta U = nC_v\Delta T \]

4. **Substitute into the work equation**:
Substituting the expressions for \( Q \) and \( \Delta U \):
\[ W = nC\Delta T - nC_v\Delta T \]
Factoring out \( n\Delta T \):
\[ W = n\Delta T \left(C - C_v\right) \]

5. **Substitute the values**:
We already calculated \( C = \frac{7R}{2} \) and for a monatomic gas, \( C_v = \frac{3R}{2} \):
\[ W = n\Delta T \left(\frac{7R}{2} - \frac{3R}{2}\right) = n\Delta T \left(\frac{4R}{2}\right) = n\Delta T \cdot 2R \]

6. **Calculate for \( n = 2 \) moles**:
Substituting \( n = 2 \):
\[ W = 2 \cdot 2R \cdot (T_2 - T_1) = 4R(T_2 - T_1) \]

### Answer for Part (b):
The work done by two moles of gas is:
\[ W = 4R(T_2 - T_1) \]

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