The vapor pressure of liquid pentane, C5H12, is 100. mm Hg at 260 K. A 0.218 g sample of liquid C5H12 is placed in a closed, evacuated 350. mL container at a temperature of 260 K. Assuming that the temperature remains constant, will all of the liquid evaporate? What will the pressure in the container be when equilibrium is reached? mm Hg

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

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