Question

6.
At 27° C temperature the
volume of o, gas is 6 litre.
At which temperature will
the volume of o, gas be
half ?
(a) -213°C
(b) -123°C
(c) -73° C
(d) -37°C​

Answer 1

Answer:-

Given:

Initial Temperature (T1) = 27° C = 273.15 + 27 = 300.15° K

Initial Volume (V1) = 6 L

Final Volume (V2) = 6/2 = 3 L (The volume is halved when the temperature is changed)

Let the Final Temperature be T2.

We know that,

Charley's law of gases states that at constant pressure the volume of a gas is directly proportional to it's temperature.

$\sf \implies \large{V = kT} \: (k \: is \: constant.)$

→ V/T = k

→ V1/T1 = V2/T2

→ 6/300.15 = 3/T2

On cross multiplication we get,

→ 6T2 = 300.15*3

→ T2 = 300.15* 3/6

→ T2 = 150° K (Approx.)

→ T2 = 150 - 273.15

→ T2 = - 123° C

Hence, the Volume halves when the temperature is decreased to - 123° C (b).