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A sample of helium gas is heated from 20.0°C to 40.0°C. This
heating process causes the gas to expand to a volume of 585
mL. What was the original volume of the helium gas?
a 292mL
b 370mL
C 548mL
d 625mL

Respuesta :

nogurt

Answer:

C. 548mL

Explanation:

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

Temperature ,T₁ = 20⁰C = 20+273.15K = 293.15K

Temperature, T₂ = 40⁰C = 40+273.15K = 313.15K

According to combined gas law,

P₁V₁/n₁T₁ = P₂V₂/n₂T₂

Where, n is moles of substance

P is pressure

P and n are considered to be a constant.

Thus, V₁/T₁ = V₂/T₂

V₁ = V₂T₁/T₂

V₁ = ?,

V₂ = 585mL = 585mL x 293.15K/313.15K

V₁ = 548mL

Hence, the original volume of the helium gas is 548mL.

What is pressure?

"Pressure is the force of all the gas particle/wall collisions divided by the area of the wall."

What is volume?

"Volume is the amount of space occupied by a substance."

What is temperature?

"Temperature is the measure of hotness or coldness expressed in terms of any of several scales, including Fahrenheit and Celsius."

Know more about Combined gas law here

https://brainly.com/question/13154969

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