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The melting point of ice 0^@C at 1 atm. ...

The melting point of ice `0^@C` at `1 atm`. At what pressure will it be `-1^@ C` ?
(Given, `V_(2)-V_(1) =(1-(1)/(0.9)) xx 10^(-3) m^(3)`).

Text Solution

Verified by Experts

Here `Delta T = (-1-0) =-1,T = 273 +0 = 273 K`
and `V_(2)-V_(1)=(1-(1)/(0.9)) xx 10^(-3) m^(3)` (given)
`L = 80 cal//g`
we have, `(Delta P)/(Delta T) = (L)/(T(V_(2)-V_(1)))`
`(Delta P)/((-1)) = (80 xx 4.2 xx 10^(3))/(273(1-(1)/(0.9))xx 10^(-3))`
`:. Delta P = 110.8 xx 10^(5) N//m^(2) = 110.8 atm`
`P_(2) -P_(1) =110.8 atm rArr P_(2) = 110.8 +P_(1) = 111.8 atm`.
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