Respuesta :
The two states of the balloon are
State 1:
pressure, p₁ = 1 atm = 101325 Pa
State 2:
volume, V₂ = 4.4 L = 4.4*0.001 m³ = 0.0044 m³
pressure, p₂ = 2.2 atm = 2.2*101325 Pa = 222915 Pa
Let V₁ = the volume in state 1.
Assume that the air in the balloon is at the same temperature at the two given states.
Then according to Boyle's Law (pV = constant), obtain
(101325 Pa)*(V₁ m³) = (222915 Pa)*(0.0044 m³)
V₁ = 0.00968 m³
= (0.00968 m³)/(0.001 m³/L) = 9.68 L
Answer: 9.68 L
State 1:
pressure, p₁ = 1 atm = 101325 Pa
State 2:
volume, V₂ = 4.4 L = 4.4*0.001 m³ = 0.0044 m³
pressure, p₂ = 2.2 atm = 2.2*101325 Pa = 222915 Pa
Let V₁ = the volume in state 1.
Assume that the air in the balloon is at the same temperature at the two given states.
Then according to Boyle's Law (pV = constant), obtain
(101325 Pa)*(V₁ m³) = (222915 Pa)*(0.0044 m³)
V₁ = 0.00968 m³
= (0.00968 m³)/(0.001 m³/L) = 9.68 L
Answer: 9.68 L
The volume of balloon at 1.0 atmis [tex]\boxed{9.68{\text{ L}}}[/tex].
Further Explanation:
A hypothetical gas comprising of a large number of randomly moving particles is called ideal gas. The collisions between such particles are considered to be perfectly elastic. Practically, no gas can be ideal so it is just a theoretical concept.
Given information:
Volume of balloon at 2.2 atm: 4.4 L
To determine:
Volume of balloon at 1.0 atm
Boyle’s law:
This law describes relationship between volume and pressure of gas. According to this law,volume of the gas is inversely proportional to its pressure, provided the temperature and the number of moles of gas remain constant. Mathematical form of Boyle’s law is,
[tex]{\text{P}} \propto \dfrac{1}{{\text{V}}}[/tex]
Or,
[tex]{\text{PV}} = {\text{k}}[/tex]
Where,
V is volume occupied by the gas.
P is the pressure of the gas.
k is a constant.
At two volumes [tex]{{\text{V}}_{\text{1}}}[/tex] and [tex]{{\text{V}}_{\text{2}}}[/tex] andpressures [tex]{{\text{P}}_{\text{1}}}[/tex] and [tex]{{\text{P}}_{\text{2}}}[/tex], equation of Boyle’s law modifies as follows:
[tex]{{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}[/tex] …… (1)
Rearrange equation (1) to calculate [tex]{{\text{V}}_{\text{2}}}[/tex].
[tex]{{\text{V}}_2} = \dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{P}}_2}}}[/tex] …… (2)
Substitute 4.4 L for [tex]{{\text{V}}_{\text{1}}}[/tex] , 2.2 atm for [tex]{{\text{P}}_{\text{1}}}[/tex] and 1.0 atm for [tex]{{\text{P}}_{\text{2}}}[/tex] in equation (2).
[tex]\begin{aligned}{{\text{V}}_2} &= \frac{{\left( {2.2{\text{ atm}}} \right)\left( {4.4{\text{ L}}} \right)}}{{\left( {{\text{1}}{\text{.0 atm}}} \right)}} \\&= 9.68{\text{ L}} \\\end{aligned}[/tex]
Learn more:
- Which statement is true for Boyle’s law: https://brainly.com/question/1158880
- Calculation of volume of gas: https://brainly.com/question/3636135
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Ideal gas equation
Keywords: Boyle’s law, P, V, k, pressure of gas, volume occupied by gas, constant, temperature, ideal gas, 2.2 atm, 1.0 atm, 4.4 L, 9.68 L.
