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stepan [7]
2 years ago
15

2. You have a 2.50 mole gas sample in a 500.0 mL flask at 25.0 °C.

Chemistry
1 answer:
SIZIF [17.4K]2 years ago
7 0

Considering the ideal gas law, the pressure of the gas sample is 122.18 atm.

<h3>What is an ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Definition of ideal gas law</h3>

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Pressure of the gas sample</h3>

In this case, you know:

  • P= ?
  • V= 500 mL= 0.5 L
  • n= 2.50 moles
  • R= 0.082 \frac{atm L}{mol K}
  • T= 25 °C= 298 K

Replacing in the ideal gas law:

P×0.5 L = 2.50 moles ×0.082 \frac{atm L}{mol K} ×298 K

Solving:

P= (2.50 moles ×0.082 \frac{atm L}{mol K} ×298 K)÷ 0.5 L

<u><em>P= 122.18 atm</em></u>

Finally, the pressure of the gas sample is 122.18 atm.

Learn more about the ideal gas law:

brainly.com/question/4147359

#SPJ1

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frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii)  ρA= 0.791 g/cm3 and,  ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

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MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

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