Question

a sample of oxygen occupies a volume of 1.6L at 91 c what will be the temperature when the volume of oxygen is reduced to 1.2l ​

Answer 1

Answer:-

Given:

Initial Volume $\sf{(V_1)}$ = 1.6 L

Initial temperature $\sf{(T_1)}$ = 91 °C + 273.15 = 364.15

Final Volume $\sf{(V_2)}$ = 1.2 L

We know that,

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

$\sf{ \implies V \: \alpha \: T}$

$\sf{v = kT(k \: is \: constant)} \\ \\ →\sf{ \frac{ V_{1} }{T _{1} } = \frac{V_{2} }{T _{2}} } \\ \\ → \sf{ \frac{1.6}{364.15} = \frac{1.2}{ T_{2}} } \\ \\ →\sf{T_{2} = \frac{364.15 \times 1.2}{1.6} } \\ \\ →\sf{T _{2} = 273.11 ° K.}$

Hence, the temperature when Volume is reduced is 273.11° K.

Answer 2

273 Kelvin

Given:

V1 = 1.6 L

V2 = 1.2 L

T1 = 91 Celsius

To Find:

T2

Solving:

To solve this type of problems we use Charles Law

The Formula is:

V1/T1 = V2/T2

Converting the temperature of Celsius to Kelvin Scale we get:

= 91 + 273

= 364 K

Substituting the values known to us in the formula given above:

1.6 L / 364 K = 1.2 L / T2

T2 = 1.2 L x 364 K / 1.6 L

T2 = 273 Kelvin

Therefore, the temperature is 273 Kelvin when the volume of Oxygen is reduced to 1.2 Litres.