Chemistry, asked by harpalvipul, 5 hours ago

An ideal gas in a sealed container has an initial volume of 2.60 L. At constant
pressure, it is cooled to 23.00 degrees Celcius, where its final volume is 1.75 L. What
was the initial temperature?

Answers

Answered by purvishah969
20

Answer:

Initial temperature was 166.77 degree Celsius

Explanation:

By Charles' Law

V1/T1=V2/T2

296*2.6/1.75= T2

T2= 439.77 K= 166.77 degree Celsius

Answered by VishnuPriya2801
52

Answer:-

Given:-

Initial volume of the gas (V₁) = 2.60 L

Final Volume of the gas (V₂) = 1.75 L

Final Temperature (T₂) = 23.00° C = 23 + 273° K = 296° K

[ ∵ °K = °C + 273 ]

We know that,

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

That is,

⟹ V ∝ T

⟹ V/T = k (constant)

So,

⟹ V₁/T₁ = V₂/T₂

Let the Initial temperature be T₁.

Hence,

⟹ (2.60)/T₁ = (1.75)/(296)

⟹ (2.60)(296) = T₁ * (1.75)

⟹ (296)(2.60) / 1.75 = T₁

⟹ 439.77° K = T₁

⟹ T₁ = 439.77 - 273°

⟹ T₁ = 166.77° C

The Initial temperature of the gas was 166.77° C.

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