(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.

Md Ariful Islam

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(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 = 14g

[(21)(1000)] / (14) = 1500cm3

T = (PV) / (n R)

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

= (9000) / (13.448)

T = 669.244k

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