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Lady bird [3.3K]
2 years ago
9

A 0.0625 tank contains 0.0925kg nitrogen at a gauge pressure of 5.17atm.Find the temperature of the gas in degree Celsius​

Physics
1 answer:
slamgirl [31]2 years ago
3 0
<h3>Answer:</h3>

-271.96 °C

<h3>Explanation:</h3>

<u>We are given;</u>

  • Volume of the tank as 0.0625 L
  • Mass of nitrogen gas as 0.0925 kg or 92.5 g
  • Pressure of the gas as 5.17 atm

we are required to calculate the temperature of the gas.

<h3>Step 1: Calculate the number of moles of nitrogen gas </h3>

Moles = Mass ÷ Molar mass

Molar mass of nitrogen gas = 28.0 g/mol

Therefore;

Moles of N₂ = 92.5 g ÷ 28.0 g/mol

                   = 3.304 moles

<h3>Step 2: Calculate the temperature of the gas;</h3>

According to the ideal gas equation;

PV = nRT , where n is the number of moles and R is the ideal gas constant, 0.082057 L.atm/mol.K

Rearranging the formula;

T = PV ÷ nR

  = ( 5.17 atm × 0.0625 L) ÷ (3.304 moles × 0.082057)

  = 1.19 K

But, °C = K - 273.15

Therefore;

T = 1.19 K - 273.15

  = -271.96 °C

Thus, the temperature of the gas will be -271.96 °C

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By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.

ANGULAR MOTION

Since the car accelerate from rest, initial velocity will be zero.

Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters

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If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.

First equation

V = U + at

Substitute all necessary parameters into the equation.

24.3 = 0 + 9.1a

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a = 2.67 m/s^{2}

Third Equation of motion

V^{2} = U^{2} + 2aS

Substitute all the necessary parameters

24.3^{2} = 0 + 2 x 2.67 x S

590.49 = 5.34S

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S = 110.58 m.

Given that the diameter of a tire is 80.2 cm,

the radius (r) will be 80.2/2 = 40.1 cm

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r = 40.1/100 = 0.401 m

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Number of revolutions = 43 rev.

Therefore, the number of revolutions the tire makes during this motion is 43 rev.

Learn more about circular motion here: brainly.com/question/6860269

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This is illustrated below:

Q = mol·ΔHv

384000 = mole of N2 x 5600

Divide both side by 5600

Mole of N2 = 384000/5600

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Next, we shall convert 68.57 moles of nitrogen, N2 to grams.

This can be obtained as follow:

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Mole of N2 = 68.57 moles.

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This can be achieved as shown below:

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