Mixture of dry air and water vapor is at temperature of 22 degree under a total pressure of 730 mm hg
Answers
Answer:
Value of t is 291 K.
Solution:- Normal boiling point is the temperature at which the pressure equals to atmospheric pressure which is 1 atm or 760 mmHg.
We are asked to calculate the temperature at which the vapor pressure is 19 mmHg.
This type of problems are solved by using Clausius Clapeyron equation:
ln(\frac{P_2}{P_1})=\frac{\Delta H_v_a_p}{R}(\frac{1}{T_1}-\frac{1}{T_2})
P_1 = 760 mmHg
T_1 = 373 K
P_2 = 19 mmHg
T_2 = t
\Delta H_v_a_p=40.67\frac{kj}{mol}
R is universal gas constant and it's value in terms of kj is 0.008134\frac{kj}{mol.K} .
Let's plug in the values in the equation and solve it:
ln(\frac{19}{760})=\frac{40.67}{0.008314}(\frac{1}{373}-\frac{1}{t})
-3.69=4891.75(0.00268-\frac{1}{t})
\frac{-3.69}{4891.75}=(0.00268-\frac{1}{t})
-0.000754=(0.00268-\frac{1}{t})
Let's keep the similar terms on same side:
\frac{1}{t}=0.00268+0.000754
\frac{1}{t} = 0.003434
t=\frac{1}{0.003434}
t = 291 K
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