InterviewSolution
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InterviewSolution
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The compressibility factor `(Z=PV//nRT)` for `N_(2)` at `223 K` and `81.06 MPa` is `1.95`, and at `373 K` and `20.265 MPa`, it is `1.10`. A certain mass of `N_(2)` occupies a volume of `1.0 dm^(3)` at `223 K` and `81.06 MPa`. Calculate the volume occupied by the same quantity of `N_(2)` at `373 K` and `20.265 MPa`.A. `3.774 dm^(3)`B. `2.77 dm^(3)`C. `5.07 dm^(3)`D. `9.30 dm^(3)` |
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Answer» Correct Answer - A For T = 223 K, P = 81.06 Mpa, Z = 1.95, and `V=1.0 dm^(3)=10^(3)cm^(3)`, we have `n=(PV)/(ZRT)=(81.06xx10^(3))/(1.95xx8.314xx223)=22.42` mol Now, at T = 373 K, P = 20,265 Mpa, Z = 1.10, the volume occupied will be `V=(ZnRT)/(P)=(1.10xx22.42xx8.314xx373)/(20.265)=3774.0 cm^(3)` `therefore V=3.774 dm^(3)` |
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