Solution to Assignment 5

# Solution to Assignment 5 - SOLUTION TO HOMEWORK#5 6.1 6.16...

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SOLUTION TO HOMEWORK #5 6.1 6.16 6.28

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6.66: The below solution is correct, but the V should be x10 -3 , not -6.

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7.12 Problem #8 and: T b T a P vap ln ln + = So, by taking the derivate of the expression of the vapor pressure with respect to temperature, we can get the answer: 22 ln ln vap vap vap vap vap a db T dP H H T dT RT dT RT H ab Ha R b R T TTR T ⎡⎤ + ⎢⎥ ΔΔ ⎣⎦ =⇒ Δ −+ = Δ = +

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Problem #9 We notice first that the water is in liquid form. Then we need to use the Poynting correction. The details are given in Illustration 7.4-7. The Poynting correction looks as follow: f L T , P () = P vap f P sat . steam exp VP P vap ( ) RT The equation simplifies to the following expression: f L T , P = f sat steam exp P vap RT and the fugacity of the saturated steam has been calculated in illustration 7.4-3, and it will not be demonstrated in here. The value is 6.7337 MPa. Now, we need to calculate the molar volume of the saturated liquid. First of all, we look in the steam tables and figure
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Solution to Assignment 5 - SOLUTION TO HOMEWORK#5 6.1 6.16...

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