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Vadim26 [7]
3 years ago
13

What pressure (in atm) will 0.44 moles of co2 exert in a 2.6 l container at 25°c?

Chemistry
1 answer:
Lilit [14]3 years ago
6 0
We can use the ideal gas law equation to find the pressure 
PV = nRTwhere 
P - pressure 
V - volume  - 2.6 x 10⁻³ m³ 
n - number of moles - 0.44 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values into the equation,
P x 2.6 x 10⁻³ m³  = 0.44 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 419 281.41 Pa
101 325 Pa is equivalent to 1 atm 
Therefore 419 281.41 Pa - 1/ 101 325 x 419 281.41 = 4.13 atm
Pressure is 4.13 atm
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An enclosed vessel contains 2.5g of 9b nitrogen and 13.3g of chlorine at s.T.P. Of What will be the partial pressure of the Il n
kow [346]

Answer:

0.535 atm

Explanation:

Since the volume of the tank is constant, we use Gay- Lussac's law to find the pressure at 180°C.

So, P₁/T₁ = P₂/T₂ where P₁ = pressure at S.T.P = 1 atm, T₁ = temperature at S.T.P = 273.15 K, P₂ = pressure of gas at 180 °C and T₂ = 180 °C = 273.15 + 180 K = 453.15 K

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P₂ = P₁T₂/T₁

Substituting the values of the variables into the equation, we have

P₂ = P₁T₂/T₁

P₂ = 1 atm × 453.15 K/273.15 K

P₂ = 1 atm × 1.66

P₂ = 1.66 atm

We now need to find the total number of moles of each gas present

number of moles of nitrogen = mass of nitrogen, m/molar mass of nitrogen molecule M

n = m/M

m = 2.5 g and M = 2 × atomic mass of nitrogen (since it is diatomic) = 2 × 14 g/mol = 28 g/mol

So, n = 2.5 g/28 g/mol

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number of moles of chlorine, n' = mass of chlorine, m'/molar mass of chlorine molecule M'

n' = m'/M'

m' = 13.3 g and M = 2 × atomic mass of chlorine (since it is diatomic) = 2 × 35.5 g/mol = 71 g/mol

So, n' = 13.3 g/71 g/mol

n' = 0.187 mol

So, the total number of moles of gas present is n" = n + n' = 0.089 mol + 0.187 mol = 0.276 mol

So, the partial pressure due to nitrogen gas, P = mole fraction of nitrogen × pressure of gas at 180 °C

P = n/n" × P₂

P = 0.089 mol/0.276 mol × 1.66 atm

P = 0.322 × 1.66 atm

P = 0.535 atm

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