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Stells [14]
3 years ago
7

if each balloon is filled with carbon dioxide gas at 20 c and 1 atmosphere calculate the mass and the number of moles of carbon

dioxide in each balloon at maxium inflation
Chemistry
1 answer:
Alona [7]3 years ago
8 0
Below is the solution:

PV = nRT 

<span>P = pressure = 1.0 atm </span>
<span>V = volume in Litres </span>
<span>n = moles CO2 = ? mol </span>
<span>R = gas constant = 0.082057 L atm mol^-1 K^-1 </span>
<span>T = temp in Kelvin = 293.15 K </span>

<span>Solve for n </span>
<span>n = PV / RT </span>

<span>Then mass CO2 = moles x molar mass </span>
<span>= n x 44.01 g/mol
</span>
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If 1.00 mol of argon is placed in a 0.500-L container at 27.0 degree C , what is the difference between the ideal pressure (as p
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Answer:

2.0 atm is the difference between the ideal pressure and  the real pressure.

Explanation:

If 1.00 mole of argon is placed in a 0.500-L container at 27.0 °C

Moles of argon = n = 1.00 mol

Volume of the container,V  = 0.500 L

Ideal pressure of the gas = P

Temperature of the gas,T = 27 °C = 300.15 K[/tex]

Using ideal gas equation:

PV=nRT

P=\frac{1.00 mol\times 0.0821 L atm/mol K\times 300.15 K}{0.500 L}=49.28 atm

Vander wall's of equation of gases:

The real pressure of the gas= p_v

For argon:

a=1.345 L^2 atm/mol^2

b=0.03219 L/mol.

(p_v+(\frac{an^2}{V^2})(V-nb)=nRT

(p_v+(\frac{(1.345 L^2 atm/mol^2)\times (1.00 mol)^2}{(0.500 L)^2})(0.500 L-1.00 mol\times 0.03219L/mol)=1.00 mol\times 0.0821 L atm/mol K\times 300.15 K

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