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?

Respuesta :

Answer:

[tex]{P_2}=2.925\ kPa[/tex]

Explanation:

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

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

Given ,  

V₁ = 10 mL

V₂ = 20 mL

P₁ = 5.85 kPa

P₂ = ?

Using above equation as:

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

[tex]{5.85\ kPa}\times {10\ mL}={P_2}\times {20\ mL}[/tex]

[tex]{P_2}=\frac{{5.85}\times {10}}{20}\ kPa[/tex]

[tex]{P_2}=2.925\ kPa[/tex]

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