Respuesta :
Answer is: 0,0030 mol of carbon dioxide.
Carbon dioxide solubility in water at 20°C and 1 atm is: 3,8·10⁻² mol/L.
Carbon dioxide solubility in water at 25°C and 1 atm is: 3,5·10⁻² mol/L.
Difference is: 0,038 mol/L - 0,035 mol/L = 0,003 mol/L.
V(carbon dioxide) = 1 L.
n(carbon dioxide) = V · c.
n(carbon dioxide) = 1 L · 0,003 mo/L.
n(carbon dioxide) = 0,0030 mol.
Carbon dioxide solubility in water at 20°C and 1 atm is: 3,8·10⁻² mol/L.
Carbon dioxide solubility in water at 25°C and 1 atm is: 3,5·10⁻² mol/L.
Difference is: 0,038 mol/L - 0,035 mol/L = 0,003 mol/L.
V(carbon dioxide) = 1 L.
n(carbon dioxide) = V · c.
n(carbon dioxide) = 1 L · 0,003 mo/L.
n(carbon dioxide) = 0,0030 mol.
[tex]\boxed{{\text{0}}{\text{.0009 mol}}}[/tex] of [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex] are released by raising the temperature of 1 L of water from [tex]{\text{20 }}^\circ {\text{C}}[/tex] to [tex]{\text{25 }}^\circ {\text{C}}[/tex].
Further Explanation:
The ideal gas equation describes the behavior of various gases under certain conditions. It shows the relationship between the pressure, temperature, volume and amount of gas.
The expression for the ideal gas equation is as follows:
[tex]{\text{PV}} = {\text{nRT}}[/tex] …… (1)
Here,
P is the pressure of the gas.
V is the volume of gas.
T is the temperature of the gas.
n is the number of moles of gas.
R is the universal gas constant.
Rearrange equation (1) to calculate the moles of gas.
[tex]{\text{n}} = \dfrac{{{\text{PV}}}}{{{\text{RT}}}}[/tex] …… (2)
The temperatures are to be converted into K. The conversion factor for this is,
[tex]{\text{0 }}^\circ {\text{C}} = {\text{273 K}}[/tex]
So the initial temperature of [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex] is calculated as follows:
[tex]\begin{aligned}{\text{Temperature}}\left( {\text{K}} \right) &= \left( {20 + 273.15} \right)\;{\text{K}}\\&= 293.15\;{\text{K}}\\\end{aligned}[/tex]
So the final temperature of [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex] is calculated as follows:
[tex]\begin{aligned}{\text{Temperature}}\left( {\text{K}} \right) &= \left( {25 + 273.15} \right)\;{\text{K}}\\&= 298.15\;{\text{K}}\\\end{aligned}[/tex]
Substitute 1 atm for P, 1 L for V, [tex]{\text{0}}{\text{.0821}}\;{\text{L}} \cdot {\text{atm/mol}} \cdot {\text{K}}[/tex] for R and 293.15 K for T in equation (2) to calculate initial moles of [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex].
[tex]\begin{aligned}{\text{n}}&= \frac{{\left( {1{\text{ atm}}} \right)\left( {{\text{1 L}}} \right)}}{{\left( {{\text{0}}{\text{.0821}}\;{\text{L}} \cdot {\text{atm/mol}} \cdot {\text{K}}} \right)\left( {293.15{\text{ K}}} \right)}}\\&={\text{0}}{\text{.416 mol}}\\\end{aligned}[/tex]
Substitute 1 atm for P, 1 L for V, [tex]{\text{0}}{\text{.0821}}\;{\text{L}} \cdot {\text{atm/mol}} \cdot {\text{K}}[/tex] for R and 298.15 K for T in equation (2) to calculate initial moles of [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex].
[tex]\begin{aligned}{\text{n}} &= \frac{{\left( {1{\text{ atm}}} \right)\left( {{\text{1 L}}} \right)}}{{\left( {{\text{0}}{\text{.0821}}\;{\text{L}} \cdot {\text{atm/mol}} \cdot {\text{K}}} \right)\left( {298.15{\text{ K}}} \right)}}\\&=0.0408{\text{mol}}\\\end{aligned}[/tex]
The released moles of [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}{\text{Released moles of C}}{{\text{O}}_{\text{2}}} &= 0.0416{\text{ mol}} - 0.0408{\text{ mol}}\\&={\text{0}}{\text{.00087 mol}}\\&\approx 0.0009{\text{ mol}}\\\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: ideal gas, pressure, volume, temperature, 293.15 K, 298.15 K, 1 atm, 1 L, CO2, initial temperature, final temperature, P, V, n, R, T, 0.00087 mol, 0.0408 mol, 0.0416 mol.
