Oxygen gas is made to undergo a process in which its molar heat capacity `C` depends on its absolute temperature `T` as `C = alpha T`. Work done by it when heated from an initial temperature `T_(0)` to a final temperature `2 T_(0)`, will be

Options

A. (a)`4alphaT_(0)^(2)`
B. (b)`(alphaT_(0)-R)(3T_(0))/(2)`
C. (c)`(3alphaT_(0)-5R)(T_(0))/(2)`
D. (d)none of these

Answer

To solve the problem, we need to determine the work done by oxygen gas when it is heated from an initial temperature \( T_0 \) to a final temperature \( 2T_0 \), given that the molar heat capacity \( C \) depends on the absolute temperature \( T \) as \( C = \alpha T \).

### Step-by-Step Solution:

1. **Understanding the relationship between heat, internal energy, and work:**
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + W \]
where \( \Delta Q \) is the heat added, \( \Delta U \) is the change in internal energy, and \( W \) is the work done by the gas.

2. **Expressing heat added (\( \Delta Q \)):**
The heat added can be expressed as:
\[ \Delta Q = nC \Delta T \]
where \( n \) is the number of moles, \( C \) is the molar heat capacity, and \( \Delta T \) is the change in temperature.

3. **Substituting for \( C \):**
Given that \( C = \alpha T \), we can write:
\[ \Delta Q = n \alpha T \Delta T \]

4. **Calculating the change in temperature (\( \Delta T \)):**
The change in temperature from \( T_0 \) to \( 2T_0 \) is:
\[ \Delta T = 2T_0 - T_0 = T_0 \]

5. **Expressing \( \Delta Q \) in terms of temperature:**
Since \( C \) depends on \( T \), we need to integrate:
\[ \Delta Q = n \int_{T_0}^{2T_0} \alpha T \, dT \]

6. **Calculating the integral:**
The integral becomes:
\[ \Delta Q = n \alpha \left[ \frac{T^2}{2} \right]_{T_0}^{2T_0} \]
Evaluating this gives:
\[ \Delta Q = n \alpha \left( \frac{(2T_0)^2}{2} - \frac{(T_0)^2}{2} \right) = n \alpha \left( \frac{4T_0^2}{2} - \frac{T_0^2}{2} \right) = n \alpha \left( \frac{3T_0^2}{2} \right) \]

7. **Calculating the change in internal energy (\( \Delta U \)):**
For diatomic gases like oxygen, the molar heat capacity at constant volume \( C_v \) is:
\[ C_v = \frac{5}{2} R \]
Thus, the change in internal energy is:
\[ \Delta U = n C_v \Delta T = n \left( \frac{5}{2} R \right) T_0 \]

8. **Substituting into the first law equation:**
Now substituting \( \Delta Q \) and \( \Delta U \) into the first law:
\[ n \alpha \left( \frac{3T_0^2}{2} \right) = n \left( \frac{5}{2} R \right) T_0 + W \]
Rearranging gives:
\[ W = n \alpha \left( \frac{3T_0^2}{2} \right) - n \left( \frac{5}{2} R \right) T_0 \]

9. **Factoring out common terms:**
\[ W = n T_0 \left( \frac{3}{2} \alpha T_0 - \frac{5}{2} R \right) \]

### Final Expression for Work Done:
Thus, the work done by the gas when heated from \( T_0 \) to \( 2T_0 \) is:
\[ W = n T_0 \left( \frac{3}{2} \alpha T_0 - \frac{5}{2} R \right) \]