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Volgvan
3 years ago
13

Why do real gases not behave exactly like ideal gases?

Chemistry
2 answers:
Korvikt [17]3 years ago
5 0

The theory assumes that collisions between gas molecules and the walls of a container are perfectly elastic, gas particles do not have any volume, and there are no repulsive or attractive forces between molecules .

BigorU [14]3 years ago
4 0

At high pressure and low temperature, real gas deviates from ideal gas. Compressibility factor is used to measure the deviation of real gas from ideal gas.

The compressibility factor is "n" which equals PV/RT.This is equal to one in case of ideal gas and this condition does not hold good for real gas.


At high pressures, the molecules in the gas tend to become closer, resulting in greater intermolecular forces. Internal molecular force holds these molecules together,prevents movement and collision and thus reduces the pressure of the gas much below than that of an ideal gas.


At low temperatures, the kinetic energy of the gas molecules also becomes less thus this results in less collision of the gas molecules and thus finally this results in the decrease in pressure of the gas molecules which would be much lower than that of an ideal gas.


For an ideal gas, the translational kinetic energy is 3/2kT. This condition does not hold good for a real gas.

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if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be​
DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
  • V is the volume of the solution. For the initial solution, V = 135\;\textbf{mL} = 0.135\;\textbf{L} for the initial solution.

n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
  • Volume of the diluted solution: 25\;\text{mL} + 135\;\text{mL}  = 160\;\textbf{mL} = 0.160\;\textbf{L}.

Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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Answer:

Option (3).

Explanation:

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