Question

The vapor pressure of liquid acetone, CH3COCH3, is 100 mm Hg at 281 K. A 6.06E-2 g sample of liquid CH3COCH3 is placed in a closed, evacuated 360. mL container at a temperature of 281 K. Calculate what the ideal gas pressure would be in the container if all of the liquid acetone evaporated.

Answer 1

Explanation:

According to ideal gas equation, product of pressure and volume equals the product of number of moles, gas constant and temperature.

Mathematically,     PV = nRT

where          P = pressure,               V = volume

                  n = no. of moles,         R = gas constant = 0.0821 atm L/mol

                   T = temperature

Since, it is known that number of moles equal mass divided by molar mass.

Hence, number of moles of given sample of acetone are as follows.

              No. of moles = $\frac{mass}{\text{molar mass of acetone}}$

                                    =  $\frac{6.06 \times 10^{-2}}{58 g/mol}$

                                    = $0.104 \times 10^{-2}$ mole

Therefore, putting the values in ideal gas equation as follows.

                         PV = nRT

$P \times 0.36 L = 0.104 \times 10^{-2} \times 0.082 atm L/mol \times 281 K$

                           = 0.359 atm

In 1 atm equal to 760 mm Hg. So, convert 0.359 atm into mm Hg as follows.

                  $0.359 atm \times \frac{760 mm Hg}{1 atm}$

                       = 272.84 mm Hg

Hence, pressure of the ideal gas will be 100 mm Hg + 272.84 mm Hg = 372.84 mm Hg

Thus, we can conclude that the ideal gas pressure in the container if all of the liquid acetone evaporated is 372.84 mm Hg.