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Alinara [238K]
3 years ago
14

A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 927

°C. The other side is evacuated and has a volume twice the size of the part containing the gas. The partition is now removed and the gas expands to fill the entire tank. Heat is now transferred to the gas until the pressure equals the initial pressure. Determine the final temperature of the gas.
Physics
1 answer:
Marianna [84]3 years ago
4 0

Answer:

The final temperature will be 3600 K.

Explanation:

Side 1

T=927°C

T=1200 K

Lets take volume of one side is V then the volume of other side will be 2V. The final volume of system will become 3 V.

The system is closed so the mass of system will remain constant .

We know that for ideal gas  P V = m R T

The final pressure is same as initial pressure so we can say that

\dfrac{T_i}{V_i}=\dfrac{T_f}{V_f}

\dfrac{1200}{V}=\dfrac{T_f}{2V+V}

T_f=3600\ K

So the final temperature will be 3600 K.

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When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels
Afina-wow [57]

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

3 0
3 years ago
A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers
Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0
3 years ago
How can a mass be hot enough to flow but still act like a solid
Lunna [17]
The inner core is solid because it is made of very dense, or heavy, materials like iron and nickel. Even though it is very hot, these materials don't
6 0
2 years ago
label the three planes and axis of movement on the figures. define each plane and axis of movement and axis of movement and give
TiliK225 [7]

Answer: it = kobe

Explanation

5 0
3 years ago
Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

Answer:

NO_{1.499}

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

n_{O} = 3.946\,kmol

Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
4 years ago
Read 2 more answers
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