An ideal gas of adiabatic exponent `gamma` is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Then, the equation of the process in terms of the variables T and V is

Options

A. `TV^(((gamma-1))/(2))=C`
B. `TV^(((gamma-2))/(2))=C`
C. `TV^(((gamma-1))/(4))=C`
D. `TV^(((gamma-2))/(4))=C`

Explanation

To solve the problem, we need to derive the equation of the process for an ideal gas when the heat transferred to the gas is equal to the decrease in its internal energy. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that the amount of heat transferred to the gas \( dQ \) is equal to the decrease in its internal energy \( -dU \). Therefore, we can write:
\[ dQ = -dU \]

2. **Applying the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ dU = dQ - dW \]
Rearranging gives:
\[ dQ = dU + dW \]
Since \( dQ = -dU \), we can substitute this into the equation:
\[ -dU = dU + dW \]
This simplifies to:
\[ 2dU + dW = 0 \]
or
\[ dW = -2dU \]

3. **Expressing Internal Energy**:
The change in internal energy for an ideal gas is given by:
\[ dU = nC_V dT \]
where \( n \) is the number of moles and \( C_V \) is the molar heat capacity at constant volume.

4. **Work Done**:
The work done on/by the gas during a volume change is given by:
\[ dW = pdV \]

5. **Substituting into the Equation**:
From the previous steps, we can express the work done as:
\[ dW = -2(nC_V dT) \]
Therefore, we have:
\[ pdV = -2nC_V dT \]

6. **Using the Ideal Gas Law**:
The ideal gas law states:
\[ pV = nRT \]
Thus, we can express pressure \( p \) as:
\[ p = \frac{nRT}{V} \]

7. **Substituting Pressure into the Work Equation**:
Substituting \( p \) into the work equation gives:
\[ \frac{nRT}{V} dV = -2nC_V dT \]
Dividing through by \( nR \) (assuming \( n \neq 0 \)):
\[ \frac{T}{V} dV = -\frac{2C_V}{R} dT \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{dV}{V} + \frac{2C_V}{R} \frac{dT}{T} = 0 \]

9. **Integrating Both Sides**:
Integrating both sides gives:
\[ \int \frac{dV}{V} + \frac{2C_V}{R} \int \frac{dT}{T} = 0 \]
This results in:
\[ \ln V + \frac{2C_V}{R} \ln T = \ln D \]
where \( D \) is a constant of integration.

10. **Exponentiating to Remove Logarithms**:
Exponentiating both sides leads to:
\[ V T^{\frac{2C_V}{R}} = D \]

11. **Final Form**:
Rearranging gives the final equation:
\[ T V^{\frac{\gamma - 1}{2}} = \text{constant} \]

### Final Answer:
The equation of the process in terms of the variables \( T \) and \( V \) is:
\[ T V^{\frac{\gamma - 1}{2}} = \text{constant} \]