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Question
A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
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Solution
For the ideal gas equation of state
PV = nRT
P1 = 500 kPa
T1 = 25°C = 25 + 273 = 298 K
P2 = 520 kPa
T2 = ?
Expansion of tyre is negligible `("V"_"constant")`
`("P"_1"V"_1)/"T"_1 = ("P"_2"V"_2)/"T"_2`
∴ T2 = `("P"_2/"P"_1)"T"_1`
= `520/500 xx 298`
= `154960/500`
= 309.92 K
T2 = 309.92 − 273
T2 = 36.9°C
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