Question

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255?

Answer 1

Hello!

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255 ?

We have the following data:

V1 (initial volume) = 1.00 L

V2 (final volume) = 1.10 L

T1 (initial temperature) = ? (in Kelvin)

T2 (final temperature) = 255 K

According to the Law of Charles and Gay-Lussac in the study of gases, in an isobaric transformation, ie when a mass under pressure maintains its constant pressure, on the other hand, as the volume increases, the temperature increases and, if the volume decreases, the temperature decreases (directly proportional to temperature and volume) . We apply the data to the formula of isobaric transformation (Charles and Gay-Lussac), we will see:

$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$

$\dfrac{1.00}{T_1} = \dfrac{1.10}{255}$

multiply the means by the extremes

$1.10*T_1 = 1.00*255$

$1.10\:T_1 = 255$

$T_1 = \dfrac{255}{1.10}$

$\boxed{\boxed{T_1 \approx 231.8\:K}}\:\:\:\:\:\:\bf\red{\checkmark}$

Answer:

The initial temperature is approximately 231.8 Kelvin

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Answer 2

Answer is 231.8 K

Explanation
We can use Charle's law to solve this problem.

Charle's law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V α T 

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 1.00 L
T₁ = ?
V₂ = 1.10 L
T₂ = 255 K (Since the unit of the given temperature is missing in the question, thought that the given temperature is in Kelvin)

By applying the formula,
1.00 L / T₁ = 1.10 L / 255 K                   
             T₁ = (1.00 L / 1.10 L) x 255 K
             T₁ = 231.8 K

Hence, the initial temperature is 231.8 K.

Assumption : the pressure of the gas is a constant.