Question

3. A sample of octane (C3H18) gas has a mass of 1.135 g at a volume of 400. cm'3 and a temperature

Answer 1

Answer:

9.65 atm

Explanation:

From the question, we are given;

• Mass of  octane is 1.135 g
• Volume of octane is 400 cm³ or 0.4 L
• Temperature of octane is 200°C

       But, K = °C + 273.15

• Therefore, the temperature is equivalent to 473.15 K

We are required to calculate the pressure of the gas.

To do this, we are going to use the ideal gas equation;

• According to the ideal gas equation;
• PV = nRT, where n is the number of moles, R is the ideal gas constant, 0.082057 L.atm/mol.K

• We can rearrange the formula to calculate the pressure;

P = nRT ÷ V

• But, n is the number of moles;

• Moles = Mass ÷ Molar mass
• Molar mass of octane is 114.23 g/mol

• Moles of octane = 1.135 g ÷ 114.23 g/mol

                                    = 0.00994 moles

Therefore;

$P = \frac{(0.00994 moles)(0.082057)(473.15)}{0.4L}$

$P=9.648atm$

P = 9.65 atm

Thus, the pressure of the gas is 9.65 atm