1.

A gas at a pressure of 1 atm is compressed at a constant pressure from 10.0 litres to 2.0 litres. If 500 J of heat energy leaves the gas calculate (i) the work done by the gas and (ii) the change in its internal energy.

Answer»

Solution :`P = 1 ATM = 101325 N//m ^(2)`
`Delta V = (2-10) =- 8` litres `=0 8 xx 10 ^(-3) m ^(3)`
Work DONE `= Delta W = P Delta V`
`= 101325 xx (-8 xx 10 ^(-3))`
`=- 826J`
Heat LEAVING the gas `= Delta Q =- 500 J`
Changle in internal energy `= Delta U =?`
`Delta U = Delta Q - Delta W`
`=- 500 + 826`
`= 326 J`


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