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jasenka [17]
2 years ago
5

The air within a piston equipped with a cylinder absorbs 565 J of heat and expands from an initial volume of 0.10 L to a final v

olume of 0.85 L against an external pressure of 1.0 atm. What is the change in internal energy of the air within the piston
Physics
1 answer:
jeyben [28]2 years ago
5 0

Answer:

489 J

Explanation:

According to the first law of thermodynamics:-

\Delta U = q + w

Where,

U is the internal energy

q is the heat

w is the work done

From the question,

q = + 565 J  (+ sign as the heat is being absorbed)

The expression for the calculation of work done is shown below as:

w=-P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

From the question,

\Delta V = 0.85 - 0.10 L = 0.75 L

P = 1.0 atm

w=-1.0\times0.75\ atmL

Also, 1 atmL = 101.3 J

So,

w=-1.0\times0.75\times 101.3\ J=-76\ J (work is done by the system)

So,

\Delta U = +565\ J-76\ J = 489\ J

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Answer:

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Explanation:

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1 year ago
A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
Mekhanik [1.2K]
The initial volume of the gas is
V_i = 600 cm^3
while its final volume is
V_f = 200 cm^3
so its variation of volume is
\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3

The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
Q=+100 J

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
\Delta U = Q-W = 100 J - (-160 J)=+260 J
where the positive sign means the internal energy of the gas has increased.
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2 years ago
Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

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Describe a theme-park ride that has constant speed but changing velocity.<br><br> Thank you!
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How about a carousel (merry go round).
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A ball is fixed to the end of a rod and is swung in a circle at a constant rate. Consider four scenarios with differing values f
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Answer:

Explanation:

the answer is found below

5 0
2 years ago
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