The temperature of a gas contained in a closed vessel increases by `1^(@)C` when pressure of the gas is increased by `1%`. The initial temperature of the gas is

Options

A. 100 K
B. `100^(@)C`
C. 250K
D. `250^(@)C`

Answer

To solve the problem, we will use the relationship between pressure and temperature for an ideal gas, which is derived from the ideal gas law. Here are the steps to find the initial temperature of the gas:

### Step 1: Understand the relationship between pressure and temperature
From the ideal gas equation \( PV = nRT \), we can derive that for a constant volume (as in a closed vessel), pressure \( P \) is directly proportional to temperature \( T \). This can be expressed as:
\[ P = kT \]
where \( k \) is a constant.

### Step 2: Set up the initial and final conditions
Let:
- Initial pressure = \( P_1 \)
- Initial temperature = \( T_1 \)
- Final pressure = \( P_2 \)
- Final temperature = \( T_2 \)

From the problem, we know:
- \( T_2 = T_1 + 1 \) (since the temperature increases by \( 1^\circ C \))
- \( P_2 = P_1 + 1\% \text{ of } P_1 = 1.01 P_1 \)

### Step 3: Write the equations for initial and final states
Using the relationship \( P = kT \):
1. For the initial state:
\[ P_1 = kT_1 \]
2. For the final state:
\[ P_2 = kT_2 \]

### Step 4: Substitute the values of \( P_2 \) and \( T_2 \)
Substituting \( P_2 \) and \( T_2 \) into the equation:
\[ P_2 = k(T_1 + 1) \]
This gives us:
\[ 1.01 P_1 = k(T_1 + 1) \]

### Step 5: Divide the equations
Now, we can divide the two equations:
\[ \frac{P_1}{P_2} = \frac{T_1}{T_2} \]
Substituting \( P_2 \) and \( T_2 \):
\[ \frac{P_1}{1.01 P_1} = \frac{T_1}{T_1 + 1} \]

### Step 6: Simplify the equation
This simplifies to:
\[ \frac{1}{1.01} = \frac{T_1}{T_1 + 1} \]

### Step 7: Cross-multiply
Cross-multiplying gives:
\[ T_1 + 1 = 1.01 T_1 \]

### Step 8: Rearrange the equation
Rearranging the equation:
\[ 1.01 T_1 - T_1 = 1 \]
\[ 0.01 T_1 = 1 \]

### Step 9: Solve for \( T_1 \)
Now, solving for \( T_1 \):
\[ T_1 = \frac{1}{0.01} = 100 \text{ Kelvin} \]

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

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