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3 years ago

Under what conditions can an ideal gas undergo a change of state without doing external work?

1 answer:

8 0

Nswer

Work done = Pressure * ΔV [ change in volume ] 

If ΔV=0, then no work is done 

Zero work thermodyanamic process is called ' isochoric process.' 

For example if a gas heated in a rigid container: the pressure and temperature 
of the gas will increase, but the volume will remain the same. 

This is called an isochoric thermodynamic process. 

It is actually possible to do work on a system without changing the 
volume, as in the case of stirring a liquid

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Oceanic water particles mainly move in circles; is this movement greater on the ocean's surface or below the surface? Explain yo

goblinko [34]

I think that the oceanic water particles mainly move in circles greater in the oceans surface because of how big the waves can be and how wind and air impact the motion. The water particles move more on the surface because of the other factors that impact it such as people, wind, air, etc...

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2 years ago

In a given chemical reaction, the energy of the products is greater than the energy of the reactants. Which statement is true fo

Flura [38]

Energy is released in the reaction

Explanation:

In a given where the energy of the products is greater than that of the reactants, we can infer that energy is released in the reaction.

This indicates that the reaction is an exothermic or exergonic reaction.

These reaction types are accompanied by release of energy.

In an exothermic change energy is released to the surroundings.
The surrounding becomes hotter at the end of the change.
This applies in exergonic reaction which leaves a reaction having more energy than it originally started with.

Learn more:

Exothermic process brainly.com/question/10567109

#learnwithBrainly

3 0

3 years ago

Read 2 more answers

A sample of helium gas has a volume of 900. milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the t

Sladkaya [172]

Answer:

The new pressure = 5.64 atm.

Explanation:

Using The general gas equation,

P₁V₁/T₁ = P₂V₂/T₂..................... Equation 1

making P₂ the subject of the equation

P₂ = P₁V₁T₂/V₂T₁ ...................... Equation 2

Where P₁ = initial pressure, V₁ = initial Volume, T₁ = Initial temperature, P₂ = final pressure, V₂ = final volume, T₂ = final Temperature.

Given: P ₁= 2.5 atm, T₁ = 298 K, V₁= 900 milliliters, T ₂= 336 K, V₂ = 450 milliliters

Substituting these values into equation 2,

P₂ = (2.5×900×336)/(298×450)

P₂ = 756000/134100

P₂ = 5.64 atm.

Thus the new pressure = 5.64 atm.

5 0

2 years ago

Help !!!! Estimate the number of breaths taken by a person during 44 years?

Lisa [10]

44 x 12. I got the 12 from the total of 12 months in a year.

44 > 40

x
12 > 10
----------
The way my teacher taught me how to estimate is look at the neighbor to 44 and 12. The only time 44 can become 50, is when the neighbor is 5 or up. Same thing for 12. Now, multiply 40 and 10.

40 x 10 = 400.

Therefore, your estimate is 400.

The real answer is 520 breaths.

6 0

2 years ago

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

7 0

3 years ago