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goblinko [34]
2 years ago
6

A 2. 75-l container filled with co2 gas at 25°c and 225 kpa pressure springs a leak. When the container is re-sealed, the pressu

re is 185 kpa and the temperature is 10°c. How many moles of gas were lost?.
Chemistry
1 answer:
Serjik [45]2 years ago
3 0

The number of moles of gas lost is  0.0213 mol. It can be solved with the help of Ideal gas law.

<h3>What is Ideal law ?</h3>

According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.

PV = nRT.

Where,

  • p = pressure
  • V = volume (1.75 L = 1.75 x 10⁻³ m³)
  • T =  absolute temperature
  • n = number of moles
  • R =  gas constant, 8.314 J*(mol-K)

Therefore, the number of moles is

n = PV / RT

State 1 :

  • T₁ = (25⁰ C = 25+273 = 298 K)
  • p₁ = 225 kPa = 225 x 10³ N/m²

State 2 :

  • T₂ = 10 C = 283 K
  • p₂ = 185 kPa = 185 x 10³ N/m²

The loss in moles of gas from state 1 to state 2 is

Δn = V/R (P₁/T₁ - P₂/T₂ )

V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N

p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)

p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)

Therefore,

Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))

    = 0.0213 mol

Hence, The number of moles of gas lost is 0.0213 mol.

Learn more about ideal gas here ;

https://brainly.in/question/641453

#SPJ1

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3 years ago
How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)
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Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide =\frac{100 g}{44 g/mol}=2.273 moles

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = 2.273\times 10^{-3} kmol

2) 1 liter of ethyl alcohol of density 0.789 g/cm^3

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = 0.789 g/cm^3

1 cm^3=1 mL

Mass of ethyl alcohol = m

m=d\times V=0.789 g/cm^3\times 1000 mL=789 g

Molar mass of  ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = \frac{789 g}{46 g/mol}=17.152 mol

17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol

3) Volume of oxygen gas,V =1.5 m^3=1500 L

1 m^3= 1000 L

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

PV=nRT

n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol

0.01632 mol = 0.01632 × 0.001 kmol=1.632\times 10^{-5} kmol

4) Volume water in mixture = 1 L

Density of water =  1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L

Mass of water = 1000 g/L\times 1 L = 1000 g

Volume of alcohol = 2.5 L

Density of alcohol =  789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L

Mass of alcohol = 789 g/L\times 2.5 L = 1972.5 g

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

\frac{1000 g}{2972.5 g}\times 100=33.64\%

Mass percentage of alcohol :

\frac{1972.5 g}{2972.5 g}\times 100=66.36\%

Moles of water :

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Moles of alcohol =

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Mole fraction of water :

\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644

Mole fraction of alcohol :

\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356

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