Question

A. At a low temperature (1 degree celsius), the volume in the cylinder was 3.8 mL. The gas pressure inside the cylinder was 764 mm Hg. Using the ideal gas equation, compute the number of moles of air. Be sure to convert the temperature and volume to appropriate units.

B. At a higher temperature, the partial pressure of air inside the cylinder was determined to be 6.32 mm Hg. Assuming that the total pressure was 764 mm Hg, determine the partial pressure of water vapor in the cylinder.

C. Relative to the volume in part A(3.8 mL), would the volume of gas inside the cylinder for part B be higher or would it be lower.

Answer 1

A:  

PV = nRT

n = (764/760)atm×0.0038L/(0.0821atm-L/K.mol × 274.15K)

= 1.697×10^-4 mol = 1.7×10^-4 mol. (Answer)

B:

P_water + P_air = 764mm Hg

P_water = 764 - 6.32

Partial pressure of water vapor = 757.68 mm Hg. (Answer)

C:

As total pressure remains same , the increase in temperature should increase the volume . So volume of cylinder should be more in part B.