contestada

If the volume of a sample of a gas is increased 5 times the original volume while the temperature is held constant, what happens to the pressure of the gas?

Respuesta :

Answer:

Therefore,

The pressure of the gas will be 1/5th of the original.

[tex]P_{2}=\dfrac{1}{5}P_{1}[/tex]

Explanation:

Given:

Let V₁ , P₁, and T₁ be the original Volume , Pressure and Temperature,

Now it Changes to V₂ , P₂ , and T₂

[tex]V_{2}=5\times V_{1}\\T_{2}=T_{1}=constant[/tex]

To Find:

What happens to the new pressure

[tex]P_{2}=?[/tex]

Solution:

Combined Gas Law:

The combined gas law combines the three gas laws:  

Boyle's Law, Charles' Law, and Gay-Lussac's Law.  

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

Hence,

[tex]\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}[/tex]

Substituting the values [tex]V_{2}=5\times V_{1}\\T_{2}=T_{1}=constant[/tex]  we get, T₁ and T₂ , V₁ will get cancelled,

[tex]P_{2}=\dfrac{1}{5}P_{1}[/tex]

That is Pressure decreases by [tex]\dfrac{1}{5}[/tex]

Therefore,

The pressure of the gas will be 1/5th of the original.

[tex]P_{2}=\dfrac{1}{5}P_{1}[/tex]