Aman heats a balloon in the oven. if the balloon initially has a volume of 0.8 liters and a temperature of 12°c, what will the volume of the balloon be after he heats it to a temperature of 300°c?

1.044 L

Explanation:

Data Given

initial volume V1 of gas in balloon = 0.8 L

initial Temperature T1 of gas in balloon = 12 °C

convert Temperature to Kelvin

T1 = °C +273

T1 = 12°C + 273 = 285 K

final Temperature T2 of gas in balloon = 100°C

convert Temperature to Kelvin

T1 = °C +273

T1 = 100°C + 273 = 373 K

final volume V2 of gas in balloon = ?

Solution:

This problem will be solved by using Charles' law equation at constant pressure.

The formula used

                    V1 / T1 = V2 / T2

As we have to find out volume, so rearrange the above equation

                    V2 = ( V1 / T1 ) x T2

Put value from the data given

                  V2 =  (0.8 L / 285 K) x 373 K

                   V2 = (0.0028 L/K) x 373 K

                   V2 = 1.044 L

So the final volume of gas in balloon= 1.044 L

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.

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