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Air is pumped into an automobile tyre's tube up to pressure pf 200kPa in the morning wheb the air temperature is `20^(@)C.`During the day the temperature rises tp `40^(@)C`and the tube expand by25. Calculate the pressure of the air in the tube at this temperature.

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Given
` P_1 = 200 Kpa`
` = 2 xx (10^5)Pa`
` T_1 = (20^@)C = 293K`
` T_2= (40^@C)= 313K`
` V_2= V_1 + 2% (V_1)`
` = 1.02(V_1)(P_2)=?`
We know
` ((P_1)(V_1)/T_1) = ((P_2)(V_2)/T_2)`
` rArr ((2 xx (10^5)V_1)/293) = ((p_2 xx 1.02 V_1)/(100 xx 313))`
` rArr P_2 = ((2 xx (10^7) xx 313)/(1.02 xx 293))`
` = 209462 Pa.`
`= 209.462KPa.`
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