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 need to use the ideal gas law and the relationship between pressure, volume, and temperature. Here's a step-by-step solution:

### Step 1: Understand the relationship between pressure, volume, and temperature
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Set up the initial conditions
Let the initial pressure be \( P \) and the initial temperature be \( T \). According to the problem, the pressure increases by 1%, which can be expressed as:
\[ P' = P + 0.01P = 1.01P \]

### Step 3: Determine the change in temperature
The problem states that the temperature of the gas increases by \( 1^\circ C \). In Kelvin, this change is:
\[ T' = T + 1 \]

### Step 4: Use the ideal gas law for initial and final states
For the initial state, we have:
\[ PV = nRT \]
For the final state, we have:
\[ P'V = nRT' \]

Substituting the values of \( P' \) and \( T' \):
\[ 1.01PV = nR(T + 1) \]

### Step 5: Divide the two equations
Dividing the final state equation by the initial state equation gives:
\[ \frac{1.01PV}{PV} = \frac{nR(T + 1)}{nRT} \]

This simplifies to:
\[ 1.01 = \frac{T + 1}{T} \]

### Step 6: Cross-multiply to solve for \( T \)
Cross-multiplying gives:
\[ 1.01T = T + 1 \]

### Step 7: Rearrange the equation
Rearranging the equation leads to:
\[ 1.01T - T = 1 \]
\[ 0.01T = 1 \]

### Step 8: Solve for \( T \)
Dividing both sides by 0.01 gives:
\[ T = \frac{1}{0.01} = 100 \text{ K} \]

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