Respuesta :
Explanation:
According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.
That is, [tex]Volume \propto Temperature[/tex]
Hence, it is given that [tex]V_{1}[/tex] is 3.50 liters, [tex]T_{1}[/tex] is 20 degree celsius, and [tex]T_{2}[/tex] is 100 degree celsius.
Therefore, calculate [tex]V_{2}[/tex] as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}[/tex]
[tex]\frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}[/tex]
[tex]V_{2}[/tex] = 17.5 liter
Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.
The volume of the gas at [tex]{\text{100 }}^\circ{\text{C}}[/tex] is [tex]\boxed{{\text{17}}{\text{.5 L}}}[/tex]
Further explanation:
Charles’s law:
Charles’s work showed that at constant pressure, the volume-temperature relationship for a fixed amount of gas is linear. In other words, Charles’s law can be stated that at constant pressure, the volume occupied by a fixed amount of a gas is directly proportional to its absolute temperature (Kelvin). This relationship is known as Charles’s law.
The mathematical representation of Charles’s law is,
[tex]{\mathbf{V}}\propto{\mathbf{T}}[/tex] [P and n are constant]
Here,
V is volume occupied by the fixed quantity of gas.
T is the temperature of a gas.
P is the pressure of a gas.
n denotes the number of moles of gas.
The relationship can also be expressed as,
[tex]\frac{{\text{V}}}{{\text{T}}}={\text{constant}}[/tex] [P and n are constant]
Or it can also be expressed as follows:
[tex]\frac{{{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}=\frac{{{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}[/tex] …… (1)
Here,
[tex]{{\text{V}}_1}[/tex] is the initial volume of gas.
[tex]{{\text{V}}_2}[/tex] is the final volume of gas.
[tex]{{\text{T}}_1}[/tex] is the initial temperature of the gas.
[tex]{{\text{T}}_2}[/tex] is the final temperature of the gas.
Rearrange equation (1) to calculate [tex]{{\text{V}}_2}[/tex].
[tex]{{\text{V}}_{\text{2}}}=\frac{{{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}[/tex] …… (2)
The value of [tex]{{\text{V}}_1}[/tex] is 3.50 L.
The value of [tex]{{\text{T}}_1}[/tex] is [tex]{\text{20 }}^\circ{\text{C}}[/tex].
The value of [tex]{{\text{T}}_2}[/tex] is [tex]{\text{100 }}^\circ{\text{C}}[/tex].
Substitute these values in equation (2).
[tex]\begin{gathered}{{\text{V}}_{\text{2}}}=\frac{{\left({{\text{3}}{\text{.50 L}}}\right)\left({{\text{100 }}^\circ{\text{C}}}\right)}}{{\left({{\text{20 }}^\circ{\text{C}}}\right)}}\\={\mathbf{17}}{\mathbf{.5 L}}\\\end{gathered}[/tex]
The volume of gas occupied at[tex]{\mathbf{100 ^\circ C}}[/tex]is 17.5 L.
Learn more:
1. Law of conservation of matter states: https://brainly.com/question/2190120
2. Calculation of volume of g as: https://brainly.com/question/3636135
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Ideal gas of equation
Keywords: Charles’s law, volume, temperature, pressure, volume temperature relationship, absolute temperature, constant pressure, relationship, V directly proportional to T, ideal gas, ideal gas equation number of moles, moles, P, n, V, T, volume of gas, 17.5 L, Kelvin, 3.50 L,
