Question

A certain quantity of a gas occupies 200 ml when collected over water at 15 degree Celsius and 745 mm pressure. It occupies 182.6 ml in the dry state at NTP. Find the vapour pressure of water at 15 degree Celsius.

Answer 1

Try to recall that at NTP, 1mole of any substance occupies 22.4L or 22400 ml of volume and also, the amount of gas at NTP will be equal to that at 15 degree Celsius and 745 mm pressure. Now, by using this you can easily answer the given question.

Complete step by step solution:
It is known to you that NTP means Normal Temperature and Pressure also known as STP (Standard Temperature and Pressure). At NTP normal temperature is 298 K and normal pressure is 1 atm=760 mm of Hg.
Calculation:
Given, pressure of moist gas, P=745 mm of Hg.
Pressure of dry gas at NTP, p= 1 atm= 760 mm of Hg.
Volume of moist gas, V=200ml
Volume of dry gas, v=182.6ml
Temperature of moist gas, T= 288K
Temperature of dry gas, t=298K
Let pressure of dry gas at 15 degree Celsius be n
Since, the amount of gas will be same in both conditions

So,\frac{{nV}}{T} = \frac{{pv}}{t} \\\ \Rightarrow \frac{{n \times 200}}{{288}} = \frac{{760 \times 182.6}}{{298}} \\\ or,n = 670.59mm{\text{ of Hg}} \\\ \end{gathered} $$ Pressure of moist gas= vapour pressure of water + pressure of dry gas $$\begin{gathered} 745 = Vapor{\text{ pressure of water + 670}}{\text{.59}} \\\ {\text{or, vapor pressure of water = 745 - 670}}{\text{.59 = 74}}{\text{.41mm of Hg}} \\\ \end{gathered} $$. **Note:** It should be remembered that vapour pressure of water is only affected by the temperature and it increases with increase in temperature. Also, you should remember that vapour pressure does not depend on the shape and size of the container and it also depends on the intermolecular force of attraction.