Equal amounts of heat are supplied to equal masses of helium and oxygen, kept at the same initial temperature. If `T_(He)andT_(O)` denote the increase in temperatures of helium and oxygen, then

Options

A. `T_(He)=T_(O)`
B. `T_(He)gtT_(O)`
C. `T_(He)ltT_(O)`
D. `T_(He)=1/16T_(O)`

Explanation

To solve the problem, we need to analyze the relationship between the heat supplied to the gases and their temperature changes, taking into account their degrees of freedom.

### Step-by-Step Solution:

1. **Understanding the Gases**:
- Helium (He) is a monoatomic gas, meaning it consists of single atoms.
- Oxygen (O2) is a diatomic gas, meaning it consists of two atoms bonded together.

2. **Degrees of Freedom**:
- The degrees of freedom for a monoatomic gas like helium is 3 (translational motion in x, y, and z directions).
- The degrees of freedom for a diatomic gas like oxygen is 5 (3 translational + 2 rotational).

3. **Heat Capacity**:
- The heat capacity at constant volume (C_v) is related to the degrees of freedom:
- For monoatomic gases: \( C_v = \frac{3}{2} R \)
- For diatomic gases: \( C_v = \frac{5}{2} R \)
- Here, \( R \) is the universal gas constant.

4. **Applying the Heat Equation**:
- The amount of heat supplied (Q) is related to the change in temperature (ΔT) and heat capacity (C_v) by the equation:
\[ Q = m \cdot C_v \cdot \Delta T \]
- Since equal masses of helium and oxygen are supplied equal amounts of heat, we can set up the equations for both gases:
- For helium:
\[ Q = m \cdot \left(\frac{3}{2} R\right) \cdot T_{He} \]
- For oxygen:
\[ Q = m \cdot \left(\frac{5}{2} R\right) \cdot T_{O} \]

5. **Equating the Heat Supplied**:
- Since both equations equal Q, we can set them equal to each other:
\[ m \cdot \left(\frac{3}{2} R\right) \cdot T_{He} = m \cdot \left(\frac{5}{2} R\right) \cdot T_{O} \]
- The mass (m) and gas constant (R) cancel out:
\[ \frac{3}{2} T_{He} = \frac{5}{2} T_{O} \]

6. **Solving for Temperature Changes**:
- Rearranging gives:
\[ T_{He} = \frac{5}{3} T_{O} \]
- This indicates that the increase in temperature of helium (T_{He}) is greater than that of oxygen (T_{O}).

7. **Conclusion**:
- Therefore, we conclude that:
\[ T_{He} > T_{O} \]

### Final Answer:
The increase in temperature of helium is greater than that of oxygen, i.e., \( T_{He} > T_{O} \).