Answer to Question #123189 in General Chemistry for Daniel

Question

(R=0.082 L atm mol-1 K-1)

i. The absolute temperature of 21 g Nitrogen gas at which gas occupies 164 mL at 6 atm.

Answer 1

(R=0.082 L atm mol-1 K-1)

21 g Nitrogen gas at which gas occupies 164 mL at 6 atm.

PV = n RT

whereby:

P = presure

V = volume

n = number of moles

R = gas constant ( 0.082 L atm mol- 1 K-1)

T = temperature

Therefore :

Nitrogen gas = 14_g

[(21)(1000)] / (14) = 1500cm³

T = (PV) / (n R)

T = [(6)(1500) ] / [ ( 0.082)(164)]

= (9000) / (13.448)

T = 669.244k