Question

A gas is enclosed in a vessel at standard temperature. At what temperature the volume of enclosed gas will be 1/6 of its volume, given that pressure remains constant. ​

Answer 1

V1 = 1 then

Because V2 = 0.1667 V1 then the formula V1/V2=t1/t2 by substitution becomes 1/0.1667 x 1 = t1/t2 or substituting t1 1/.01667 x 1 = 296.15/t2

Solving for t2 then becomes

t2 = 296.15 x .1667

or t2 = 49.42 K or about -223 C

It the gas is air then it is likely to not be totally gas by this point.

Answer 2

Answer:

⇒ PV = nRT

Here P, n and R are constants.

Therefore V ∝ T

Let V1 be V and V2 be V/6 initial T = 273.15 K

∴ V1/V2 = 273.15/T2

∴ T2 = 273.15/6 = 45.53 K

Hence, the volume will be 1/6 of its original volume at 45.53 K