Question

In a closed container, the vapor pressure of 10 mL of ethanol at 20 degrees C is 5.85 kPa. What is the vapor pressure of 20 mL of ethanol at the same temperature?

Answer 1

`{P_2}=2.925\ kPa`

Step-by-step explanation:

At same temperature and same number of moles, Using Boyle's law

`{P_1}* {V_1}={P_2}* {V_2}`

Given ,

V₁ = 10 mL

V₂ = 20 mL

P₁ = 5.85 kPa

P₂ = ?

Using above equation as:

`{P_1}* {V_1}={P_2}* {V_2}`

`{5.85\ kPa}* {10\ mL}={P_2}* {20\ mL}`

`{P_2}=\frac{{5.85}* {10}}{20}\ kPa`

`{P_2}=2.925\ kPa`