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. `4 alpha T_(0)`
B. `(alpha T_(0) - R) (3 T_(0))/(2)`
C. `(3 alpha T_(0) - 5 R) (T_(0))/(2)`
D. none of these

Explanation

To solve the problem of finding the work done by oxygen gas when it is heated from an initial temperature \( T_0 \) to a final temperature \( 2T_0 \) with a molar heat capacity \( C = \alpha T \), we can follow these steps:

### Step 1: Understand the relationship between heat, molar heat capacity, and temperature change
The heat \( Q \) added to the gas can be expressed as:
\[ Q = nC\Delta T \]
where \( n \) is the number of moles, \( C \) is the molar heat capacity, and \( \Delta T \) is the change in temperature.

### Step 2: Substitute the expression for molar heat capacity
Given that \( C = \alpha T \), we can substitute this into the equation for heat:
\[ Q = n(\alpha T)\Delta T \]

### Step 3: Determine the change in temperature
The change in temperature \( \Delta T \) from \( T_0 \) to \( 2T_0 \) is:
\[ \Delta T = 2T_0 - T_0 = T_0 \]

### Step 4: Substitute \( \Delta T \) into the heat equation
Now, substituting \( \Delta T \) into the equation for heat:
\[ Q = n\alpha T (T_0) \]

### Step 5: Express \( Q \) in terms of integration
Since \( C = \alpha T \) varies with temperature, we need to integrate to find the total heat added:
\[ Q = n\int_{T_0}^{2T_0} \alpha T \, dT \]

### Step 6: Perform the integration
Integrating \( Q \):
\[ Q = n\alpha \left[ \frac{T^2}{2} \right]_{T_0}^{2T_0} = n\alpha \left( \frac{(2T_0)^2}{2} - \frac{(T_0)^2}{2} \right) \]
\[ = n\alpha \left( \frac{4T_0^2}{2} - \frac{T_0^2}{2} \right) = n\alpha \left( \frac{3T_0^2}{2} \right) \]

### Step 7: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
where \( \Delta U \) is the change in internal energy and \( W \) is the work done by the gas.

### Step 8: Calculate the change in internal energy for an ideal gas
For an ideal gas, the change in internal energy \( \Delta U \) can be expressed as:
\[ \Delta U = nC_V\Delta T \]
For oxygen, \( C_V = \frac{5}{2}R \), thus:
\[ \Delta U = n\left(\frac{5}{2}R\right)(T_0) = \frac{5}{2}nRT_0 \]

### Step 9: Substitute \( Q \) and \( \Delta U \) into the first law equation
Substituting into the first law:
\[ n\alpha \left( \frac{3T_0^2}{2} \right) = \frac{5}{2}nRT_0 + W \]

### Step 10: Solve for work \( W \)
Rearranging gives:
\[ W = n\alpha \left( \frac{3T_0^2}{2} \right) - \frac{5}{2}nRT_0 \]

### Step 11: Factor out common terms
Factoring out \( nT_0 \):
\[ W = nT_0 \left( \frac{3\alpha T_0}{2} - \frac{5R}{2} \right) \]

### Final Expression
The work done by the gas when heated from \( T_0 \) to \( 2T_0 \) is:
\[ W = nT_0 \left( \frac{3\alpha T_0 - 5R}{2} \right) \]