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 type at afternoon ?

Explanation

For ideal gas equation of state
`PV=nRT`
`P_(1)=500kPa, T_(1)=25^(@)C=25+273=298K, P_(2)=520kPa, T_(2)=?`
Expansion of tyre is negligible `(V_("constant"))`
`(P_(V)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`therefore" "T_(2)=((P_(2))/(P_(1)))T_(1)=(520)/(500)xx298=(154960)/(500)-=309.92K`
`T_(2)=309.92-273, T_(2)=36.9^(@)C`