Respuesta :
Data Given:
Temperature = 273 K
Pressure = 1 atm
Mole of O₂ = 1 g / 32 g.mol⁻¹ = 0.03125 mol
Mole of H₂ = 1 g / 2.016 g.mol⁻¹ = 0.0496 mol
Mole of Ar = 1 g / 39.94 g.mol⁻¹ = 0.025 mol
Solution:
As Ideal Gas Equation is given as,
P V = n R T
Solving for V,
V = n R T / P
Solving for O₂;
V = (0.03125 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm
V = 0.699 L
Solving for H₂;
V = (0.0496 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm
V = 1.11 L
Solving for Ar;
V = (0.025 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm
V = 0.55 L
Result:
The correct answer is H₂. 1 g of H₂ occupies more volume as compare to 1 g of O₂ and 1 g of Ar.
Temperature = 273 K
Pressure = 1 atm
Mole of O₂ = 1 g / 32 g.mol⁻¹ = 0.03125 mol
Mole of H₂ = 1 g / 2.016 g.mol⁻¹ = 0.0496 mol
Mole of Ar = 1 g / 39.94 g.mol⁻¹ = 0.025 mol
Solution:
As Ideal Gas Equation is given as,
P V = n R T
Solving for V,
V = n R T / P
Solving for O₂;
V = (0.03125 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm
V = 0.699 L
Solving for H₂;
V = (0.0496 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm
V = 1.11 L
Solving for Ar;
V = (0.025 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm
V = 0.55 L
Result:
The correct answer is H₂. 1 g of H₂ occupies more volume as compare to 1 g of O₂ and 1 g of Ar.
The ideal behaviour of the gas states the volume of the gas particles is not considered compared to the total volume of the container. The intermolecular forces and interactions of the gas particles are thought to be negligible.
1 gm of [tex]\rm H_{2}[/tex] will have more volume at STP.
How to calculate the volume?
Given,
- Pressure = 1 atm
- Temperature = 273 k
Calculate the moles of [tex]\rm O_{2}[/tex] (n):
[tex]\begin{aligned}\rm n &= \dfrac{ 1 \;\rm g }{32 \;\rm gmol^{-1} }\\\\&= 0.03125 \;\rm mol\end{aligned}[/tex]
Calculate the moles of [tex]\rm H_{2}[/tex] (n):
[tex]\begin{aligned}\rm n &= \dfrac{1 \;\rm g}{2.016 \;\rm gmol^{-1}}\\\\&= 0.0496 \;\rm mol\end{aligned}[/tex]
Calculate the moles of Ar (n):
[tex]\begin{aligned}\rm n &= \dfrac{1 \;\rm g}{39.94 \;\rm gmol^{-1}} \\\\&= 0.025 \;\rm mol\end{aligned}[/tex]
The ideal gas equation is given as:
[tex]\begin{aligned}\rm P V &= \rm n R T\\\\\rm V &= \dfrac{\rm n R T}{\rm P}\end{aligned}[/tex]
Calculating the volume for Oxygen:
[tex]\begin{aligned}\rm V &= \dfrac{(0.03125 \;\rm mol \times 0.08205 \;\rm atm\;L\;mol^{-1}K^{-1}\times 273 \;\rm K) }{1 \;\rm atm}\\\\\rm V &= 0.699 \;\rm L\end{aligned}[/tex]
Calculating the volume for Hydrogen:
[tex]\begin{aligned}\rm V &= \dfrac{ (0.0496\;\rm mol \times 0.08205 \;\rm atm\;\rm L\;\rm mol^{-1}K^{-1} \times 273 \;\rm K)}{1 \;\rm atm}\\\\\rm V &= 1.11 \;\rm L\end{aligned}[/tex]
Calculating the volume for Argon:
[tex]\begin{aligned} \rm V &= \dfrac{(0.025 \;\rm mol \times 0.08205 \;\rm atm\;\rm L\;r\m mol^{-1}K^{-1}\times 273 \;\rm K) }{1 \;\rm atm}\\\\\rm V &= 0.55 \;\rm L\end{aligned}[/tex]
Therefore, option B is correct.
Learn about ideal gas here:
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