A gas at 400 K and 10.0 L
expands to 20.0 L. What
is its new temperature in
Kelvin?​


Sagot :

Given:

[tex]V_{1} = \text{10.0 L}[/tex]

[tex]T_{1} = \text{400 K}[/tex]

[tex]V_{2} = \text{20.0 L}[/tex]

Required:

[tex]T_{2}[/tex]

Strategy:

This is a gas law problem. What gas law should we use?

Since the given quantities are volume and temperature, we will use Charles' law. According to this gas law, the volume occupied by a gas is directly proportional to its absolute temperature keeping the pressure and the amount of gas constant.

Caution: The temperature must be converted to kelvin. If the given temperature is in degree Celsius, add 273 or 273.15 to convert it to kelvin.

The formula used for Charles' law is

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

where:

[tex]V_{1} = \text{initial volume}[/tex]

[tex]T_{1} = \text{initial temperature}[/tex]

[tex]V_{2} = \text{final volume}[/tex]

[tex]T_{2} = \text{final temperature}[/tex]

Solution:

Starting with the formula of Charles' law

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

Multiplying both sides of the equation by T₁T₂ to eliminate the denominators

[tex]V_{1}T_{2} = V_{2}T_{1}[/tex]

Dividing both sides of the equation by V₁ to solve for T₂

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

Substituting the given values and solving for V₂

[tex]T_{2} = \text{400 K} \times \dfrac{\text{20.0 L}}{\text{10.0 L}}[/tex]

Therefore, the final temperature in kelvin is

[tex]\boxed{T_{2} = \text{800 K}}[/tex]

Answer:

T₂ = 800 K

[tex]\\[/tex]

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