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Question
Air is pumped into an automobile tyre's tube up to a pressure of 200 kPa in the morning when the air temperature is 20°C. During the day the temperature rises to 40°C and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature.
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Solution
Here,
\[ P_1 = 2 \times {10}^5 Pa \]
P2 = ?
T1 = 293 K
T2 = 313 K
V2 = V1 + 0.02 V1 = V1 (1.02)
Now,
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
\[\Rightarrow \frac{2 \times {10}^5 V_1}{293} = \frac{P_2 V_1 (1.02)}{313} \]
\[ \Rightarrow P_2 = \frac{2 \times {10}^5 \times 313}{293 \times 1.02} = 209 \text{ kPa }\]
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