Answer :
Answer:
The right answer to this question is no, all of the liquid will not evaporate, there will be 8.61 ×10⁻⁴ moles or 6.22×10⁻² grams left in the container
At equilibrium the pressure in the container will be the vapor pressure of liquid pentane which is = 100. mm Hg
Explanation:
To solve this we list the known values as follows
vapor pressure of liquid pentane = 100 mmHg = 13.33 KPa
Temperature T = 260 K
Volume of container = 350 mL = 0.00035 m³
The number of moles of liquid pentane = n
The universal gas constant = R = 8.314 J/(mol·K)
Thus From the ideal gas equation PV = nRT →
Thus plugging in the values in the above equateion we have
n = [tex]\frac{PV}{RT} = \frac{(13330)(0.00035)}{(8.314)(260)}[/tex] = 2.16×10⁻³ moles
Hence the number of moles in 0.218 g sample of liquid pentane C₅H₁₂ with molar mass = 72.15 g/mol = 0.218/72.15 = 3.02×10⁻³ moles
Hence the number of moles present in the sample placed in the closed evacuated container = 3.02×10⁻³ moles
However number of moles to completely evaporate at 100 mmHg and 260 K is 2.16×10⁻³ moles hence, 3.02×10⁻³ moles - 2.16×10⁻³ moles, or 8.61 ×10⁻⁴ moles will be left in the container
converting the value in moles to mass we have number of moles, n = mass/(molar mass)
Therefore the mass = number of moles × molar mass = 8.61 ×10⁻⁴ × 72.15 = 6.22 × 10⁻² grams left in the container
The pressure in the container at equilibrium will be vapor pressure of liquid pentane C₅H₁₂, or 100. mm Hg