An ideal gas is expanded so that the amount of heat given is equal to the decrease in internal energy of the gas. The gas undergoes the process `PV^((6)/(5))=` constant. The gas may be

Options

A. `He`
B. `O_(2)`
C. `Ar`
D. all of the above

Answer

To solve the problem, we start with the information given:

1. **Understanding the Process**: We have an ideal gas that is expanded such that the heat given (dQ) is equal to the decrease in internal energy (dU). Mathematically, this can be expressed as:
\[ dQ = -dU \]

2. **Using the First Law of Thermodynamics**: According to the first law of thermodynamics:
\[ dQ = dU + dW \]
where \(dW\) is the work done by the system. Since \(dQ = -dU\), we can rewrite this as:
\[ -dU = dU + dW \]
This leads to:
\[ dW = -2dU \]

3. **Understanding the Polytropic Process**: The problem states that the gas undergoes a process defined by:
\[ PV^{\frac{6}{5}} = \text{constant} \]
In a polytropic process, the relationship between pressure (P), volume (V), and the polytropic index (n) is given by:
\[ PV^n = \text{constant} \]
Here, \(n = \frac{6}{5}\).

4. **Relating Heat Capacities**: The molar heat capacity \(C\) for a polytropic process is given by:
\[ C = C_v + \frac{R}{1 - n} \]
where \(C_v\) is the molar heat capacity at constant volume and \(R\) is the universal gas constant.

5. **Finding the Value of \(C\)**: Since we know that \(dQ = -dU\), we can express \(C\) in terms of internal energy. For an ideal gas, the change in internal energy is:
\[ dU = C_v dT \]
Therefore, we can substitute this into our equation for \(dQ\):
\[ dQ = -C_v dT \]
Thus, we have:
\[ C = -C_v \]

6. **Substituting Values**: From the relationship of heat capacities in a polytropic process:
\[ -C_v = C_v + \frac{R}{1 - n} \]
Rearranging gives:
\[ -2C_v = \frac{R}{1 - n} \]
Substituting \(n = \frac{6}{5}\):
\[ -2C_v = \frac{R}{1 - \frac{6}{5}} = \frac{R}{-\frac{1}{5}} = -5R \]
Therefore:
\[ C_v = \frac{5R}{2} \]

7. **Finding \(\gamma\)**: The ratio of heat capacities \(\gamma\) is defined as:
\[ \gamma = \frac{C_p}{C_v} \]
For an ideal gas, \(C_p = C_v + R\). Thus:
\[ C_p = \frac{5R}{2} + R = \frac{7R}{2} \]
Therefore:
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{7R}{2}}{\frac{5R}{2}} = \frac{7}{5} = 1.4 \]

8. **Identifying the Gas**: The value of \(\gamma = 1.4\) corresponds to a diatomic gas (like oxygen or nitrogen).

Thus, the gas may be a **diatomic gas**.