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)`).

The melting point of ice `0^@C` at `1 atm`. At what pressure will it b

1.	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)`).
Answer» 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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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)`).

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