Question

Determine the volume (in L) of a closed container with a pressure of 6.88 kPa containing 1.93×10^-1 g of C₄H₆ at a temperature of -112.92 °C. Report your answer to ***((( three significant figures )))

Answer 1

The volume can be calculated by the formula

V = nRT/P

1 kPa equals to 0.00986923267 Atm

Therefore 6.88 kPa = 0.00986923267 x 6.88

= 0.0679 atm

Temperature has to be converted to Celsius to Kelvin

T = -112.92 + 273 = 160.08 K

n is the number of moles

It is calculated by

n = 1.93 x 10^-1g or 0.193 g

Molecular weight of C₄H₆= 54.0916 g/mol

= (0.193 g x 1 mol)/54.0916 g

= 0.003568 mol

The volume can be calculated by the formula

V = nRT/P

R = 0.0821 L.atm/mol.K

V = (0.003568 mol)(0.0821 L.atm/mol.K)(160.08 K)/0.0679 atm

Volume = 0.6906 L

The result has to be expressed in three significant figures then

Volume = 0.691 L