The temperature of `5 mol` of gas which was held at constant volume was change from `100^(@)C` to `120^(@)C`. The change in internal energy was found to have `80 J`. The total heat capacity of the gas at constant volume will be equal to

Options

A. `8 JK^(-1)`
B. `0.8 JK^(-1)`
C. `4 JK^(-1)`
D. `0.4 JK^(-1)`

Explanation

To find the total heat capacity of the gas at constant volume, we can follow these steps:

### Step 1: Understand the Problem
We are given:
- Number of moles of gas (n) = 5 mol
- Initial temperature (T1) = 100°C
- Final temperature (T2) = 120°C
- Change in internal energy (ΔU) = 80 J

We need to find the total heat capacity at constant volume (C_v).

### Step 2: Convert Temperatures
Since the change in temperature can be calculated in Celsius or Kelvin (the difference is the same), we can find the change in temperature (ΔT):
\[ \Delta T = T2 - T1 = 120°C - 100°C = 20°C \]
This is also equivalent to:
\[ \Delta T = 20 K \]

### Step 3: Apply the First Law of Thermodynamics
At constant volume, the first law of thermodynamics states:
\[ \Delta Q = \Delta U + W \]
Where W (work done) is zero at constant volume. Thus:
\[ \Delta Q = \Delta U \]
This means that the heat added to the system (ΔQ) is equal to the change in internal energy (ΔU):
\[ \Delta Q = 80 J \]

### Step 4: Relate Heat Capacity to Internal Energy Change
The change in internal energy can also be expressed in terms of heat capacity at constant volume:
\[ \Delta U = n C_v \Delta T \]
Substituting the known values:
\[ 80 J = 5 \, \text{mol} \times C_v \times 20 K \]

### Step 5: Solve for C_v
Rearranging the equation to solve for \(C_v\):
\[ C_v = \frac{80 J}{5 \, \text{mol} \times 20 K} \]
Calculating this gives:
\[ C_v = \frac{80}{100} = 0.8 \, \text{J/(mol K)} \]

### Step 6: Calculate Total Heat Capacity
The total heat capacity at constant volume (C_total) for 5 moles of gas is:
\[ C_{total} = n \times C_v = 5 \, \text{mol} \times 0.8 \, \text{J/(mol K)} = 4 \, \text{J/K} \]

### Final Answer
The total heat capacity of the gas at constant volume is:
\[ \boxed{4 \, \text{J/K}} \]