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3 years ago

0.235 moles of a diatomic gas expands doing 205 J of work while the temperature drops 88 K. Find Q.? (Unit=J)

Physics

1 answer:

Harman [31]3 years ago

5 0

The heat released by the gas is -225 J

Explanation:

First of all, we have to calculate the change in internal energy of the gas, which for a diatomic gas is given by

$\Delta U = \frac{5}{2}nR\Delta T$

where

n = 0.235 mol is the number of moles

$R=8.314\cdot J/mol K$ is the gas constant

$\Delta T = -88 K$ is the change in temperature

Substituting,

$\Delta U = \frac{5}{2}(0.235)(8.314)(-88)=-430 J$

Now we can us the 1st law of thermodynamics to find the heat absorbed/released by the gas:

$\Delta U = Q -W$

where

$\Delta U = -430 J$ is the change in internal energy

Q is the heat

W = 205 J is the heat done by the gas

Solving for Q,

$Q=\Delta U + W = -430 + 205 =-225 J$

Since the sign is negative, it means the heat has been released by the gas.

Learn more about thermodynamics:

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Answer:

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3 years ago

Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same

marissa [1.9K]

Answer:

maximum static frictional force of the wall on the book (Increasing)

normal force of the wall on the book (Decreasing)

weight of the book (Not changing)

Explanation:

Now according to Newton's third law of motion

"Every action has equal but opposite reaction"

By the data given in question, Albert was pressing the book against the wall.Now, Albert started to reduce his force up against the wall.

First we have to consider all the forces applied on book in this scenario.

1. Weight of book acting downwards (y-axis)

2. Friction between book and wall acting upward (y-axis)

3. Albert's force on book against wall (x-axis)

4. Normal reaction of wall against Albert's force (x-axis)

Now, when Albert reduced his force, new scenario will be

1. Weight will be remain constant as it is W = mg

Neither mass nor acceleration due to gravity changed, so weight acting upon the book will remain same.

2. When Albert reduced force, normal reaction of wall reduced against it according to Newton's third law of motion

3. Now notice that friction is a force which acts in accordance with the applied force. For example if a box is placed at floor, no friction is applied, but when you drag the box, friction starts to act and increases until its limit comes. So, when Albert reduced his force, weight will try to pull the book and maximum static friction will increase to hinder the movement of book downwards.

Notice that maximum static friction will hinder the book from movement, since Albert reduced his force, but wight didn't pull the book, which means that maximum static friction increased to hinder downward motion.

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3 years ago

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3 years ago

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The frequency of a wave is 560 Hz. What is it’s period

Crazy boy [7]

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3 years ago

What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Taya2010 [7]

Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:

Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Answer:

The Force act on each side is 2.43×10⁴N

Explanation:

Given data

n=3 mol

L=0.3 m

Temperature=20.0°C=293 K

To find

Force F

Solution

To get force act on each side it would employ by

F=P.A

Where P is pressure

A is Area

First we need to find pressure by applying ideal gas law

$P.V=nRT\\P=\frac{nRT}{V}\\ P=\frac{(3mol)(8.315J/mol)(293K)}{(0.3m*0.3m*0.3m)}\\P=27.069*10^{4}Pa$

So The Force is given as:

$F=P.A\\F=(27.069*10^{4} )(0.3m*0.3m)\\F=2.43*10^{4}N$

The Force act on each side is 2.43×10⁴N

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