The molar enthalpy of vaporization of boron tribromide is 30.5 kj/mol, and its normal boiling point is 91°c. what is the vapor p
1 answer:
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Answer:
20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.
Explanation:
Heat is being consumed during vaporization and heat is being released during condensation.
To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.
Molar mass of water = 18.02 g/mol
Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.
So, to vaporize 9.00 g of water, of heat or 20.3 kJ of heat is being consumed
As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.
Option B: high pressure and low temperature
A gas is more soluble under high pressure and low temperature conditions.
On increasing temperature of a gas, its kinetic energy increases. The increase in kinetic energy increases the motion of particles of gas this causes most of the gaseous particles to escape from the gas phase. Thus, less particles are available to dissolve in liquid and solubility decreases.
The effect of pressure on solubility of gas can be explained with the help of Henry's law. According to the law, at constant temperature, solubility of gas and partial pressure of gas are related to each other as follows:
Here, p is the partial pressure of the gas, is Henry's law constant, and
c is the concentrate of the gas.
According to above relation, concentration of gas decreases on decreasing partial pressure. Thus, on increasing pressure, concentration of gas increases this increases the solubility of gas in liquid.
Therefore, solubility of gas is greatest at high pressure and low temperature.