Question

LR Terences 1 pt TUTOR Partial Pressures 1 pt A sample of a gas mixture contains the following quantities of three gases. 1 pt 1 pt compound СО CO2 mass 1.32 g 3.14 g 3.27 g 1 pt 1 pt 1 pt The sample has: 1 pt volume = 2.50 L temperature = 16.6 °C What is the partial pressure for each gas, in mmHg? What is the total pressure in the flask? 1 pt 1 pt CO mmHg 1 pt CO2 mmHg 1 pt SF6 mmHg 1 pt total mmHg 1 pt Submit Show Approach Show Tutor Steps 1 pt

Answer 1

From the ideal gas equation

Partial pressure of CO

= moles of CO x R x Temperature / volume

= (mass of CO / molecular weight of CO) x (R x Temperature / volume)

= (1.32 g / 28 g/mol) x [0.0821 L-atm/mol-K x (16.6+273)K / 2.50L]

= 0.44835 atm x (760 mmHg / atm)

= 340.75 mmHg

Similarly,

Partial pressure of CO2

= moles of CO2 x R x Temperature / volume

= (mass of CO2 / molecular weight of CO2) x (R x Temperature / volume)

= (3.14 g / 44 g/mol) x [0.0821 L-atm/mol-K x (16.6+273)K / 2.50L]

= 0.6787 atm x (760 mmHg / atm)

= 515.81 mmHg

Similarly,

Partial pressure of SF6

= moles of SF6 x R x Temperature / volume

= (mass of SF6 / molecular weight of SF6) x (R x Temperature / volume)

= (3.27 g / 146 g/mol) x [0.0821 L-atm/mol-K x (16.6+273)K / 2.50L]

= 0.213 atm x (760 mmHg / atm)

= 161.89 mmHg

Total pressure = Partial pressure of CO + Partial pressure of CO2 + Partial pressure of SF6

= 340.75 + 515.81 + 161.89

= 1018.45 mmHg