Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the air temperture is `22^(@)C` During the day , temperature ries to `42^(@)C` and the tube expands by 2% a The pressure of the air in the tube at this temperature, will be approximately

Options

A. 212 kPa
B. 209 kPa
C. 206 kPa
D. 200 kPa

Explanation

To find the pressure of the air in the automobile tube at the new temperature, we can use the ideal gas law and the relationship between pressure, volume, and temperature. The steps are as follows:

### Step 1: Convert Temperatures to Kelvin
- The initial temperature \( T_1 \) is given as \( 22^\circ C \).
- Convert to Kelvin:
\[ T_1 = 22 + 273 = 295 \text{ K} \]
- The final temperature \( T_2 \) is given as \( 42^\circ C \).
- Convert to Kelvin:
\[ T_2 = 42 + 273 = 315 \text{ K} \]

### Step 2: Determine the Initial Pressure and Volume
- The initial pressure \( P_1 \) is given as \( 200 \text{ kPa} \).
- Let the initial volume \( V_1 \) be \( V \).

### Step 3: Calculate the Final Volume
- The tube expands by 2%, so the final volume \( V_2 \) can be calculated as:
\[ V_2 = V_1 + 0.02 V_1 = 1.02 V_1 = 1.02 V \]

### Step 4: Apply the Ideal Gas Law
Using the ideal gas law, we can relate the initial and final states of the gas:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Rearranging gives us:
\[ P_2 = P_1 \cdot \frac{V_1}{V_2} \cdot \frac{T_2}{T_1} \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
- \( P_1 = 200 \text{ kPa} \)
- \( V_2 = 1.02 V \)
- \( T_1 = 295 \text{ K} \)
- \( T_2 = 315 \text{ K} \)

This gives:
\[ P_2 = 200 \cdot \frac{V}{1.02 V} \cdot \frac{315}{295} \]
The \( V \) cancels out:
\[ P_2 = 200 \cdot \frac{1}{1.02} \cdot \frac{315}{295} \]

### Step 6: Calculate the Values
Now calculate \( P_2 \):
1. Calculate \( \frac{1}{1.02} \approx 0.9804 \)
2. Calculate \( \frac{315}{295} \approx 1.0678 \)

Now substitute these values:
\[ P_2 \approx 200 \cdot 0.9804 \cdot 1.0678 \]
Calculating this:
\[ P_2 \approx 200 \cdot 1.046 \approx 209.2 \text{ kPa} \]

### Final Answer
The pressure of the air in the tube at the new temperature will be approximately \( 209.2 \text{ kPa} \).

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