Answer: No, a<span>t high pressures, volume of a real gas does not compare with the volume of an ideal gas under the same conditions.
Reason:
For an ideal gas, there should not be any intermolecular forces of interaction. However, for real gases there are intermolecular forces of interaction like dipole-dipole and dipole-induced dipole. Further, at high pressures, molecules are close by. Hence, extend of these intermolecular forces is expected to be high. This results in decreases in volume of real gas. Thus, </span>volume of a real gas does not compare with the volume of an ideal gas under the same conditions.
The volume of Helium gas needed for storage is 2.00 L (answer C)
<u><em> calculation</em></u>
The volume of Helium is calculated using ideal gas equation
That is Pv =nRT
where;
P( pressure) = 203 KPa
V(volume)=?
n(number of moles) = 0.122 moles
R(gas constant) = 8.314 L.Kpa/mol.K
T(temperature)= 401 K
make V the subject of the formula by diving both side by P
V=nRT/p
V={[0.122 moles x 8.314 L. KPa/mol.K x 401 K] / 203 KPa} = 2.00 L
Answer:
Mass = 0.697 g
Explanation:
Given data:
Volume of hydrogen = 1.36 L
Mass of ammonia produced = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
First of all we will calculate the number of moles of hydrogen:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K
1.36 atm.L = n × 22.43 atm.L/mol
n = 1.36 atm.L / 22.43 atm.L/mol
n = 0.061 mol
Now we will compare the moles of hydrogen and ammonia:
H₂ : NH₃
3 : 2
0.061 : 2/3×0.061 = 0.041
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.041 mol × 17 g/mol
Mass = 0.697 g
Potassium or any other metals.
