The temperature of a gas contain in a closed vessel increased by `2^(0)C` when the pressure is increased by `2%` the intial temperature of the gas is

Options

A. 200K
B. 100K
C. `200^(@)C`
D. `100^(@)C`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We are given that the temperature of a gas in a closed vessel increases by \(2^{\circ}C\) when the pressure is increased by \(2\%\). We need to find the initial temperature of the gas.

### Step 2: Set up the initial and final conditions
Let:
- Initial pressure = \(P_0\)
- Final pressure = \(P_0 + 0.02 P_0 = 1.02 P_0\)
- Initial temperature = \(T_0\)
- Final temperature = \(T_0 + 2\)

### Step 3: Use the Ideal Gas Law
According to the Ideal Gas Law, for a constant volume, the ratio of pressure to temperature remains constant:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Substituting our values:
\[ \frac{P_0}{T_0} = \frac{1.02 P_0}{T_0 + 2} \]

### Step 4: Simplify the equation
We can cancel \(P_0\) from both sides (since \(P_0 \neq 0\)):
\[ \frac{1}{T_0} = \frac{1.02}{T_0 + 2} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ T_0 + 2 = 1.02 T_0 \]

### Step 6: Rearrange the equation
Rearranging the equation, we get:
\[ 1.02 T_0 - T_0 = 2 \]
\[ 0.02 T_0 = 2 \]

### Step 7: Solve for \(T_0\)
Dividing both sides by \(0.02\):
\[ T_0 = \frac{2}{0.02} = 100 \text{ Kelvin} \]

### Conclusion
The initial temperature of the gas is \(100 \text{ Kelvin}\).

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