Question

Using the van der waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 c is____________atm. (a= 6.49L^2-atm/mol^2, b=0.0562 L/mol)

A. 1.50
b. 0.676
c. 0.993
d. 1.48
e. 1.91

Answer 1

D. 1.48atm

Step-by-step explanation:

Van der waals equation is given as:

(P +an²/v²) (v - nb) = nRT

Where;

P = pressure (atm)

V = volume (L)

R = gas constant (0.0821 Latm/molK)

a and b = gas constant specific to each gas

T = temperature (K)

n = number of moles

According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol

Hence;

(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15

(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638

(P + 0.0291) (22.316) = 33.638

22.316P + 0.649 = 33.638

22.316P = 33.638 - 0.649

22.316P = 32.989

P = 32.989/22.316

P = 1.478

P = 1.48atm