We first consider the gases that will be present in that sample.
First, there will be nitrogen, as stated. Second, there will also be water in the form of water vapor. For this, we need the vapor pressure of water at 23.0 °C, which is about 21.0 mmHg. Now, the sum of the vapor pressures of the gases will be equivalent to the total pressure. So the pressure of nitrogen gas is:
785 - 21
= 764 mmHg
Answer:
Explanation:
1)
Given data:
Initial volume of balloon = 0.8 L
Initial temperature = 12°C ( 12+273= 285 K)
Final temperature = 300°C (300+273 = 573 K)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 0.8 L .573 K / 285 K
V₂ = 458.4 L / 285
V₂ = 1.61 L
2)
Initial pressure = 204 kpa
Initial temperature = 29°C ( 29 + 273 = 302 K)
Final temperature = ?
Final pressure = 300 kpa
Solution:
P₁/T₁ = P₂/T₂
T₂ = T₁P₂/P₁
T₂ = 302 K . 300 kpa / 204 kpa
T₂ = 90600 K/ 204
T₂ = 444.12 K
3)
Given data:
Initial volume = 14 L
Initial pressure = 2.1 atm
Initial temperature = 100 K
Final temperature = 450 K
Final volume = ?
Final pressure = 1.2 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm
V₂ = 13230 L / 120
V₂ = 110.25 L
Answer:
Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L
Explanation:
More the pressure of the gas, more will be its solubility.
So, for two different pressure, the relation between them is shown below as:-
Given ,
P₁ = 1 atm
P₂ = 0.76 atm
C₁ = 6.8 × 10⁻⁴ mol/L
C₂ = ?
Using above equation as:
<u>Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L</u>
ADD THEM all, and then divide by four. Thats what I would do!
