10 points and brailiest to correct answer plz

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion

stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 240 K

= initial volume of gas =

= final volume of gas =

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0

3 years ago

You place ammonium nitrate crystals in water and stirred. As you do so, the container becomes cold to the touch. This an example

ankoles [38]

The process of flask becoming cold is due to endothermic reaction.

Answer: Option B

<u>Explanation:</u>

So two kinds of heat transfer can be possible in any chemical reaction. If the sample is considered as system and the sample container is considered as the surrounding, then heat transfer can occur between them.

If the heat is transferred from the surrounding to the system , then it is an endothermic reaction. And in those cases, the sample holder will be becoming colder. This is because the heat from the surrounding that is the container will be utilized to complete the reaction.

While when there is transfer of heat from the system to surrounding , it will be exothermic reaction and the beaker will be getting hot in this process. So in the present case, the container is becoming cold because of occurrence of endothermic process.

8 0

4 years ago

When the displacement of a mass on a spring is 12 a the half of the amplitude, what fraction of the mechanical energy is kinetic

Sveta_85 [38]

You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).

And you know that the potential energy, PE, is [ 1/2 ] k (x^2)

Then, use x = A, to calculate the PE in the point where ME = PE.

ME = PE = [1/2] k (A)^2.

At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2

=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME

So, if PE is 1/4 of ME, KE is 3/4 of ME.

And the answer is 3/4

7 0

4 years ago

Your body also transforms what energy

natita [175]

Chemical energy is transformed in your body

4 0

3 years ago

What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.

Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:

$|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82$

4 0

4 years ago