1 mol of an ideal gas undergoes reversible isothermal expansion from an initial volume V, to a final volume, `10 V_(1) ` and does 10 kJ of work. The initial pressure was `1 xx10^(7) Pa. a`) Calculate `V_(1),` b) if there were 2 mol of gas, what must have been its temperature?

Options

A. `2.21 xx 10^(-4) , 120.13 K`
B. `4.34 xx 10 ^(-4) , 261. 13 K`
C. `1.01 xx 10 ^(-4) , 60. 06 K`
D. `3.21 xx 10 ^(-4) , 161 . 13 K`

Explanation

(a) For reversible isothermal expansion , `w = - 2. 303 RT log "" (V _(2))/( V _(1)) = 2. 303 n RT log "" ( P _(1))/( P _(2))`
At constant `T, p _(1) V _(1) = p _(2) V _(2).` therefore `1 xx 10 ^(7) xx V _(1) = p _(2) xx 10 V _(1) implies p _(2) = 10 ^(6) Pa`
Substituting value of presure in expansion for work , we have
`- 10 xx 10 ^(3) = - 2. 303 xx 1 xx 8. 314 xx T log "" ( 10 ^(7))/(10 ^(6)) `
Solving we get `T = 522. 27 K,` Substituting in the gas equation `pV = nRT.`
`1 xx 10 ^(7) xx V _(1) = 1 xx 8. 314 xx 522. 27 or V _(1) = 4.34 xx 10 ^(-4)`
(b) For 2 mol, ` T= ( 522. 27)/(2) = 261. 13 K`