Solveeit Logo
Login

Question

Question: If \[{{50c}}{{{m}}^3}\] of hydrogen is collected over water at \({{1}}{{{7}}^ \circ }{{C}}\) and \({...

If 50cm3{{50c}}{{{m}}^3} of hydrogen is collected over water at 17∘C{{1}}{{{7}}^ \circ }{{C}} and 750Hg{{750Hg}}pressure. Calculate the volume of dry gas at STP. The water vapor pressure at 17∘C{{1}}{{{7}}^ \circ }{{C}} is 14mmHg{{14mmHg}}.

Explanation

Solution

Total pressure of gas and vapor pressure of water is given. From these values, we can determine the value of pressure of dry hydrogen gas. Ideal gas law can be used to find the volume of dry hydrogen gas at STP condition.

Complete step by step answer:
It is given that vapor pressure of water, P=14mmHg{{P = 14mmHg}}
Initial pressure can be calculated by subtracting the water pressure from the given pressure since total pressure is the sum of initial pressure and vapor pressure of water.
i.e. P1=750−14mmHg=736mmHg{{{P}}_1} = {{750 - 14mmHg = 736mmHg}}
Initial Volume of the hydrogen, V1=50cm3{{{V}}_1} = 50{{c}}{{{m}}^3}
Initial temperature, T1=17∘C=273+17=290K{{{T}}_1} = {{1}}{{{7}}^ \circ }{{C = 273 + 17}} = 290{{K}}
As we know that at STP condition, the temperature is 0∘C{0^ \circ }{{C}} and pressure is 760mmHg760{{mmHg}}.
Final pressure, P2=760mmHg{{{P}}_2} = 760{{mmHg}}
Final temperature, T2=0∘C=273K{{{T}}_2} = {0^ \circ }{{C = 273K}}
Now we can use the ideal gas equation.
i.e. P1V1T1=P2V2T2\dfrac{{{{{P}}_1}{{{V}}_1}}}{{{{{T}}_1}}} = \dfrac{{{{{P}}_2}{{{V}}_2}}}{{{{{T}}_2}}}
On substituting the values, we get
736×50290=760×V2273\dfrac{{736 \times 50}}{{290}} = \dfrac{{760 \times {{{V}}_2}}}{{273}}
Thus the final volume can be calculated by:
V2=736×50×273290×760=10046400220400=45.58cm3{{{V}}_2} = \dfrac{{736 \times 50 \times 273}}{{290 \times 760}} = \dfrac{{10046400}}{{220400}} = 45.58{{c}}{{{m}}^3}

Thus, we can say that the volume of dry gas at STP is 45.58cm345.58{{c}}{{{m}}^3}.

Additional information:
When a gas molecule is heated, the particles in it move faster and gain energy. This has an effect on the walls of containers and thus pressure is increased. Similarly, when the molecule is cooled, the reverse will happen. This is explained in Gay-Lussac’s law. It is also known as pressure-temperature law.

Note: When temperature is changed, vapor pressure of water is also changed. Thus we can say that both of them are directly proportional. When gas is collected over water, the total pressure can be calculated by adding pressure of gas and the vapor pressure of water.