Determine how many liters 8.47 g of carbon dioxide gas would occupy at the following conditions. 211 K and 149 kPa

Answer 1

Given:

mass of CO2 = 8.47 g

temperature = 211 K

pressure = 149 kPa

Asked for:

volume in liters

Ideal gas law:

PV = nRT

P = 149 kPa x 1 atm / 101.325 kPa = 1.47 atm

V = ?

n = moles = 8.47 g CO2 x 1 mol CO2/44 g = 0.1925 moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 211K

solving for V, we have...

V = nRT/P = (0.1925)(0.0821)(211) / 1.47

V = 2.27 L

If you wanted to use R in units of kPa, we could do the problem this way...

V = nRT/P = (0.1925 mol)(8.314 LkPa/Kmol)(211) / 149 kPa = 2.27 L