Question

A given mass of gas occupies \[919\text{ }mL\] volume in dry state at $STP$ Same mass of gas collected over water at \[15{}^\circ C\text{ }\And \text{ }750\text{ }mm\] of pressure occupy volume of $1L$ Calculate vapor pressure of water at \[15{}^\circ C\]

Answer 1

Hint: We know that at \[NTP\] $1$ mole of any substance occupy \[22400\text{ }ml\] of volume as well as also amount of gas of \[NTP\] which will be equal to $15$ degree Celsius as well as \[745\text{ }mm\] pressure. Thus by using this we can easily answer given questions. The formula for gas equation is given by:
\[\dfrac{{{P}_{1}}\times {{V}_{1}}}{{{T}_{1}}}\text{ }=\text{ }\dfrac{{{P}_{2}}\times {{V}_{2}}}{{{T}_{2}}}\]

Complete step-by-step answer:
We know that \[NTP\] means Normal Temperature and Pressure which is also known as $STP$ that is Standard Temperature and Pressure. At given \[NTP\] normal temperature is \[298\text{ }K\] along with normal pressure which is \[1\text{ }atm\text{ }=\text{ }760\text{ }mm\text{ }of\text{ }Hg\]
Thus, given pressure of the moist gas \[P\text{ }=\text{ }760\text{ }mm\text{ }of\text{ }Hg\]
Pressure of the dry gas \[NTP\text{ }{{P}_{total}}\text{ }=\text{ }760\text{ }mm\text{ }of\text{ }Hg\]
Volume of the moist gas \[{{V}_{1}}\text{ }=\text{ }919ml\]
Volume of the dry gas \[{{V}_{2}}\text{ }=\text{ }100ml\]
Temperature of the moist gas \[{{T}_{1}}\text{ }=\text{ }273K\]
Temperature of the dry gas \[{{T}_{2}}\text{ }=\text{ }273\text{ }+\text{ }15\text{ }=\text{ }288K\]
Let pressure of the dry gas be $15$ degree Celsius and since amount of the gases will be of same in both condition and with using gas equation;
\[\dfrac{{{P}_{1}}\times {{V}_{1}}}{{{T}_{1}}}\text{ }=\text{ }\dfrac{{{P}_{2}}\times {{V}_{2}}}{{{T}_{2}}}\]
By substituting the values we get;
\[\Rightarrow \dfrac{760\times 919}{273}\text{ }=\text{ }\dfrac{{{P}_{2}}\times 1000}{288}\]
\[\Rightarrow {{P}_{2}}\text{ }=\text{ }\left[ \left( \dfrac{760\times 919}{273} \right)\times \left( \dfrac{288}{1000} \right) \right]\]
\[\Rightarrow {{P}_{2}}\text{ }=\text{ }736.82mmHg\]

Therefore, we know that \[{{P}_{dry\text{ }gas}}\text{ }=\text{ }{{P}_{total}}\text{ }-\text{ }{{P}_{water\text{ }vapor}}\]
\[{{P}_{dry\text{ }gas}}\text{ }=\text{ }{{P}_{total}}\text{ }-\text{ }{{P}_{water\text{ }vapor}}\]
 From here we need to find vapor pressure of water which is given by
\[\Rightarrow {{P}_{water\text{ }vapor}}\text{ }=\text{ }{{P}_{total}}\text{ }-\text{ }{{P}_{dry\text{ }gas}}\]
\[\Rightarrow {{P}_{water\text{ }vapor}}\text{ }=\text{ }750\text{ }-\text{ }736.82\]
\[\Rightarrow {{P}_{water\text{ }vapor}}\text{ }=\text{ }13.18\text{ }mm\]

Therefore, vapor pressure of water at \[15{}^\circ C\text{ }is\text{ }13.18mm\]

Note: Note that we should be remembering that the vapour pressure of water is only affected by temperature as well as it increases with an increase in the temperature. Along with that we should remember vapour pressure doesn't depend on shape as well as size of container which also depend on intermolecular force of attraction.