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The melting point of ice is 0^(@)C at 1 atm. At what pressure it will be -1^(@)C ? |
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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))` `thereforeDeltaP=132xx10^(5)N//m^(2)=132atmorP_(2)-P_(1)=132atm`. `thereforeP_(2)=132+P_(1)=133atm`. |
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