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1. A sample of gas has a volume of 6.20 L at 20°C at a pressure of 1.10 atm. What is its volume at the same temperature and at a pressure of 0.925 atm? (Hint: What equation will you use if the temperature is constant?).

Respuesta :

Answer:

7.37 L is the new volume

This can be understood as a system that maintains its constant T ° and if the moles of the gas are not modified, the pressure varies inversely proportional to the volume.

Explanation:

We can apply the Ideal Gases Law, to solve this problem. Moles are the same in both cases, the T° also keeps on constant.

P . V = n . R . T      For this first situation

P . V = n . R . T      For the second situation

V = n . R . T / P

n . R . T will be the same, because n is the number of moles, T means T° and R is a physic constant, so for the first case we can define:

P₁ . V₁, and for the second case, P₂ . V₂

We match both equations, and we replace data

6.20L. 1.10 atm = 0.925 atm . V₂

(6.20L. 1.10 atm) / 0.925 atm = V₂

V₂ = 7.31 L

If the volume decreases, the pressure increases more because the molecules collide more with the walls of the container but the opposite happens where if the volume increases (as in the case of exercise), the pressure will be lower. Gas molecules will not collide so much with the walls

Lanuel

The volume of the gas at the same temperature is equal to 7.37 Liters.

Given the following data:

  • Initial volume = 6.20 L
  • Initial pressure = 1.10 atm.
  • Final pressure = 0.925 atm
  • Temperature = 20°C

To find the volume of the gas at the same temperature, we would apply Boyle's law:

Mathematically, Boyle's law is given by the formula;

[tex]P_1V_1 = P_2V_2[/tex]

Where;

  • [tex]P_1[/tex] is the original (initial) pressure.
  • [tex]P_2[/tex] is the final pressure.
  • [tex]V_1[/tex] is the original (initial) volume.
  • [tex]V_2[/tex] is the final volume.

Substituting the given parameters into the formula, we have;

[tex]6.20 \times 1.10 = 0.925 \times V_2\\\\6.82 = 0.925V_2\\\\V_2=\frac{6.82}{ 0.925} \\\\V_2= 7.37 \;L[/tex]

Final volume = 7.37 Liters.

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