Saved Bookmarks
| 1. |
The melting point of ice is 0^(0)C at 1 atm. At what pressure will it be -1^(0)C ? |
|
Answer» Solution :Here `DeltaT = (- 1 - 0) = - 1, T = 273 + 0 = 273K and V_(2) - V_(1) = (1-1/(0.9))xx10^(-3)m^(3)` (given) L = 80 `cal//g` : We have,`(DELTAP)/(DeltaT)=L/(T(V_(2)-V_(1))) (or) (DeltaP)/(-1)=(80xx4.2xx10^(3))/(273(1-1/0.9)xx10^(-3))` `THEREFORE DeltaP = 132 xx10^(5) N//m^(2)=132 "atm" (or) P_(2)-P_(1)=132"atm" thereforeP_(2)=132 +P_(1)=133"atm"` |
|