What pressure in kPa is exerted by 94.60 g CO2 at 57.00 ℃ in 0.7500 L?

7859.98 KPa

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of CO2 = 94.60g

Temperature (T) = 57℃

Volume (V) = 0.75 L

Pressure (P) =?

Step 2:

Determination of the number of mole of CO2.

Mass of CO2 = 94.60g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Number of mole of CO2 =?

Number of mole = Mass/Molar Mass

Number of mole of CO2 = 94.60/44

Number of mole of CO2 = 2.15 moles

Step 3:

Conversion of celsius temperature to Kelvin temperature.

Temperature (Kelvin) = temperature (celsius) + 273

Temperature = 57℃ + 273 = 330K

Step 4:

Determination of the pressure. This is illustrated below:

The pressure can be obtained by using the ideal gas equation as shown below:

PV = nRT

Temperature (T) = 330K

Volume (V) = 0.75 L

Number of mole (n) = 2.15 moles

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =?

PV = nRT

0.75 x P = 2.15 x 0.082 x 330

Divide both side by 0.75

P = (2.15 x 0.082 x 330)/0.75

P = 77.572 atm

Step 5:

Conversion of the pressure in atm to KPa. This is illustrated below:

1 atm = 101.325 KPa

Therefore, 77.572 atm = 77.572x101.325 = 7859.98 KPa

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