A jar contains a gas and a few drops of water. The pressure in the jar is 830mmof mercury, The temperature of the jar is reduced by 1%. The saturation vapour pressures of water at the two temperatures are 30mm of mercury and 25mm of mercury. Calculate the new pressure in the jar.

Explanation

At temperature T_(1), the total pressure is 830mm of mercury. Out of this, 30mm if mercury is due to the vapour and 800mm of murcury is due to te gas. As thetemperature decreases. The pressure due to the gas decreases according to the gas law. Here the volume is constant, so,
`(p_(2))/(T_(2))=(p_(1))/T_(1))`
`or, p_(2)=(T-(2))/(T_(1))p_(1).`
As t_(2) is 1% less than T_(1).`
`T_(2)=0.99T_(1)`
and hence,
`p_(2)=0.99p_(1)`
`=0.99xx800mm of mercury =729mm of mercury.`
The vapour is still saturated and hence, its oressure is 25mm of mercury. The total pressure at the reduced temperature is
`p=(792+25)mm of mercury`
=817 mm of mercury.