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The melting point of ice is 0^(0) C at 1 atm. At what pressure it will be- 1^(0) C? |
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Answer» SOLUTION :Here `Delta = (-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)))` or `(Delta P)/((- 1)) = (80 xx 4.2 xx 10^(3))/(273 (1 - (1)/(0.9) ) xx 10^(-3))` `therefore Delta P = 132 xx 10^(5) N //m^(2) = 132 ` atm or `P_(2) - P_(1) = 132` atm `therefore P_(2) = 132 + P_(1) = 133` atm. |
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