Question

The vapour pressure of water at 300 k in a closed container is 0.4 atm . If the volume of container is doubled , its vapour pressure at 300 k will be

Answer 1

Vapour pressure at 300 K will be 0.2 atm

Explanation:

To calculate the new vapor pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

$P_1V_1=P_2V_2$     (at constant temperature and constant number of moles)

where,

$P_1$ = initial pressure of gas= 0.4 atm

$V_1$ =  initial volume of gas = v

$P_2$ = final pressure of gas= ?

$V_2$ = final volume of gas = 2v

Putting values in above equation, we get:

$0.4\times v=P_2\times 2v$

$P_2=0.2atm$

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