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?

Answer 1

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