Answer to Question #138458 in General Chemistry for Javier

Question

A 125 mL bubble of hot gases at 215 ∘C and 1.74 atm escapes from an active volcano.

What is the new volume of the bubble outside the volcano where the temperature is -16 ∘C and the pressure is 0.64 atm?

Answer 1

Given that P1= 1.74 atm = 1.74atm*760mmHg/1atm = 1322.4mmHg

T1= 215°C = 215+273 = 488k

V1= 125L

and P2= 0.64 atm = 0.64atm*760mmHg/1atm= 486.4mmHg

T2 = -16°C = -16+273= 257k

Since, P1V1/T1 = P2V2/T2 therefore;

1322.4mmHg*125L/488k= 486.4mmHg*n/257k

n = 1322.4mmHg*125L*257k/488k/486.4mmHg

n = V2 =178.975L

The new volume of the gas is therefore 178.975L