An ideal gas with the adiabatic exponent `gamma` undergoes a process in which its internal energy relates to the volume as `U=aV^alpha`, where a and `alpha` are constants. Find the work performed by the gas and the amount of heat to be transferred to this gas to increase its internal energy by `DeltaU`

Answer

To solve the problem, we need to find the work performed by the gas and the amount of heat transferred to the gas to increase its internal energy by \(\Delta U\).

### Step-by-Step Solution:

1. **Understand the Internal Energy Relation**:
The internal energy \( U \) of the gas is given by the relation:
\[ U = aV^\alpha \]
where \( a \) and \( \alpha \) are constants.

2. **Relate Internal Energy to Temperature**:
For an ideal gas, the internal energy can also be expressed as:
\[ U = n C_V T \]
where \( n \) is the number of moles, \( C_V \) is the molar specific heat at constant volume, and \( T \) is the temperature.

3. **Use the Ideal Gas Law**:
From the ideal gas law, we have:
\[ PV = nRT \]
Rearranging gives:
\[ T = \frac{PV}{nR} \]

4. **Substituting Temperature into Internal Energy**:
Substitute \( T \) into the internal energy equation:
\[ U = n C_V \left( \frac{PV}{nR} \right) = \frac{C_V PV}{R} \]

5. **Equating the Two Expressions for Internal Energy**:
Set the two expressions for \( U \) equal to each other:
\[ \frac{C_V PV}{R} = aV^\alpha \]
Rearranging gives:
\[ P = \frac{aR}{C_V} V^{\alpha - 1} \]

6. **Finding Work Done by the Gas**:
The work done \( W \) by the gas during a volume change can be expressed as:
\[ W = \int P \, dV \]
Substitute \( P \) into the work integral:
\[ W = \int \frac{aR}{C_V} V^{\alpha - 1} \, dV \]
This integrates to:
\[ W = \frac{aR}{C_V} \cdot \frac{V^\alpha}{\alpha} \bigg|_{V_0}^{V_f} \]
Therefore, we can express it as:
\[ W = \frac{aR}{C_V \alpha} (V_f^\alpha - V_0^\alpha) \]

7. **Relating Work to Change in Internal Energy**:
Since \( U = aV^\alpha \), the change in internal energy \( \Delta U \) is:
\[ \Delta U = a(V_f^\alpha - V_0^\alpha) \]
Thus, we can relate work done to change in internal energy:
\[ W = \frac{R}{C_V \alpha} \Delta U \]

8. **Finding Heat Transfer**:
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Rearranging gives:
\[ Q = \Delta U + W \]
Substituting for \( W \):
\[ Q = \Delta U + \frac{R}{C_V \alpha} \Delta U = \Delta U \left( 1 + \frac{R}{C_V \alpha} \right) \]

### Final Results:
- **Work Done**:
\[ W = \frac{R}{C_V \alpha} \Delta U \]
- **Heat Transferred**:
\[ Q = \Delta U \left( 1 + \frac{R}{C_V \alpha} \right) \]