Using ln(Psat) = A − 5000/T (Psat in atm, T in K), estimate the saturation pressure of water at 50 °C (assume A is calibrated to water so that this form is accurate near 50 °C). Choose the closest value.

Difficulty: Medium

Correct Answer: 0.13

Explanation:


Introduction / Context:
Compact Clausius–Clapeyron-type expressions are widely used to estimate saturation pressures near a reference temperature. Here, ln(Psat) = A − 5000/T represents an integrated form with constants chosen for water over a limited range. The task is to estimate Psat at 50 °C and pick the closest option.


Given Data / Assumptions:

  • T = 50 °C = 323 K.
  • Equation form: ln(Psat) = A − 5000/T, with A previously fitted for water in this range.
  • Answers are approximate (two decimal places in atm).


Concept / Approach:
Although A is not explicitly provided, standard steam-table data state that the saturation pressure of water at 50 °C is approximately 12.3 kPa, which equals about 0.121–0.123 atm. Among the discrete options, the best match is 0.13 atm. This is consistent with the expectation from a CC-type fit over modest temperature intervals.


Step-by-Step Solution (Reasoning by Reference Values):

Recall from data: at 40 °C, Psat ≈ 0.073 atm; at 60 °C, Psat ≈ 0.199 atm.Interpolate qualitatively: at 50 °C, Psat should lie between ≈0.07 and ≈0.20 atm, nearer to 0.12 atm.Compare choices: 0.11 and 0.13 are the nearest; 0.13 is closer to 0.121–0.123 atm.Select 0.13 atm as the closest among the given options.


Verification / Alternative check:
Using ln(Psat) = A − 5000/323 and calibrating A with any one known point (e.g., at 373 K, Psat = 1 atm) yields an A value that, when used at 323 K, also produces ≈0.12 atm, consistent with the selection.


Why Other Options Are Wrong:

  • 0.07 and 0.09 are characteristic of lower temperatures (≈40–45 °C).
  • 0.11 underestimates the known 50 °C pressure.
  • 0.20 is closer to ≈60 °C.


Common Pitfalls:
Expecting exactness from simplified correlations; always pick the closest value and remember that constants like A are fit-dependent.


Final Answer:
0.13

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