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Suppose you have a piston in a cylinder where the gas temperature is 26.0° C and has a volume of 1.75 L. What is the volume of the gas if you move the assembly to a cooler region with a temperature of 3.00° C? Assume the gas pressure within the cylinder is constant. Express your answer in L.

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skyluke89

Answer:

1.62 L

Explanation:

Charle's law for ideal gases states that for a gas kept at constant pressure, the ratio between volume and temperature is constant:

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where in this problem we have

[tex]V_1 = 1.75 L[/tex] is the initial volume of the gas

[tex]T_1 = 26.0^{\circ}+273=299 K[/tex] is the initial temperature of the gas

[tex]V_2[/tex] is the final volume of the gas

[tex]T_2 = 3.0^{\circ}+273=276 K[/tex] is the final temperature

Solving the equation for V2, we find

[tex]V_2 = \frac{V_1}{T_1} T_2 = \frac{1.75 L}{299 K}(276 K)=1.62 L[/tex]

oladayoademosu

The volume of the gas if you move the assembly to a cooler region with a temperature of 3.00° C is 1.62 L

Charles Law

For a given mass of gas at constant pressure, the volume, V is directly proportion to the temperature in kelvin, T.

So, V ∝T

V₁/T₁ = V₂/T₂

Given that the gas has an initial temperature of 26.0 °C and has a volume of 1.75 L, we need to find its volume when its moved to a cooler region of temperature 3.00 °C.

So,

  • V₁ = initial volume of gas = 1.75 L,
  • T₁ = initial temperature of gas = 26.0 °C = 26.0 + 273 = 299 K,
  • V₂ = final volume of gas and
  • T₂ = final temperature of gas = 3.00 °C = 3.00 + 273 = 276 K

Volume of gas at 3.00 °C

Making V₂ subject of the formula, we have

V₂ = V₁T₂/T₁

Substituting the values of the variables into the equation, we have

V₂ = V₁T₂/T₁

V₂ = 1.75 L × 276 K/299 K

V₂ = 483 LK/299 K

V₂ = 1.62 L

So, the volume of the gas if you move the assembly to a cooler region with a temperature of 3.00° C is 1.62 L

Learn more about Charles' law here:

https://brainly.com/question/22814353

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