1. A sample of helium gas is heated from 20.0°C to 40.0°C. This heating
process causes the gas to exland to a volume of 585 mL. What was
the original volume of the helium gas?

V₂ = 530.5 mL

Explanation:

Given data:

Initial temperature = 20.0°C

Final temperature = 40.0 °C

Final volume = 585 mL

Initial volume = ?

Solution:

Initial temperature = 20.0°C (20+273 = 293 K)

Final temperature = 40.0 °C (40+273 = 323 K)

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂  

V₂ = 585 mL × 293 K / 323 K

V₂ = 171405 mL.K / 323 K

V₂ = 530.5 mL

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.