`2.8 xx 10^(-2) kg ` of nitrogen is expanded isothermally and reversibly at 300 k from 15.15 `xx 10^(5) Nm^(-2)` when the work done is found to be - 17. 33 kJ .Find the final pressure .

Answer

Final pressure `=1456.8 Nm^(-2)`

Explanation

Given : Mass of nitrogen `= W= 2.8 xx 10^(-2) kg`
Temperature = T = 300 K
Work obtained in expansion `=W_(max)= - 17 .33 kJ =- 17330 J`
Initial pressure `=P_(1) = 15. 15 xx 10^(5) Nm^(-2) = 1.515 xx 10^(6) Nm^(-2)`
Molar mass of nitrogen `(N_(2)) = M_(N_(2)) = 28 xx 10^(-3) kg mol^(-1)`
Final pressure `= P_(2) = ?`
Number of moles of `N_(2) = n. (W)/(M_(N_(2))) = (2.8 xx 10^(-2))/(28xx 10^(-3)) = 1 mol`
`W_(max) = - 2.303 xx nRT log _(10) .(P_(1))/(P_(2))`
`17330 = 2.303 xx 1 xx 8.314 xx 300 xx log _(10). (1.515 xx 10^(6))/(P_(2))`
`:. (17330)/(2.303 xx 1xx 8.314 xx 300) = [ log_(10) 1.515 xx 10^(6) - log_(10) P_(2)]`
`3.017 = 6 . 1804 - log _(10) P_(2)`
`:. log_(10) P_(2) = 6.1804 - 3.017 =3 . 1634`
`:. P_(2) = `Antilog 3.1634
`= 1456.8 Nm^(-2)`