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MrRissso [65]
2 years ago
14

An ideal gas initially is allowed to expand isothermally until its volume of 1.6 L and pressure is 5 kPa, undergoes isothermal e

xpansion until its volume is 8 L and its pressure is 1 kPa.
1. Calculate the work done by the gas. Answer in units of kJ.
2. Find the heat added to the gas during this process. Answer in units of kJ.
Physics
1 answer:
Marina86 [1]2 years ago
7 0

(a) The work done by the gas during the isothermal expansion is -25.6 J.

(b) The heat added to the gas during this process is 25.6 J.

<h3>Net work done by the ideal gas against the external pressure</h3>

The net work done by the ideal gas in the isothermal expansion is calculated as follows;

W(net) = ΔP x ΔV

W(net) = ( 1 kPa - 5 kPa) x (8L - 1.6 L)

W(net) = -25.6 kPa.L

W(net) = -25.6 J

<h3>Head added to the gas</h3>

The heat added to the gas is calculated as follows;

W = -Q

-25.6 J = -Q

Q = 25.6 J

Learn more about Isothermal expansion here: brainly.com/question/17192821

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Romashka-Z-Leto [24]

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v=at\implies a=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}t

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x=\dfrac12at^2\implies a=\dfrac{2(49\,\mathrm m)}{t^2}

So we have

\dfrac{30\,\frac{\mathrm m}{\mathrm s}}t=\dfrac{2(49\,\mathrm m)}{t^2}\implies t=3.3\,\mathrm s

and the answer is C.

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3 years ago
Do you think any significant discoveries have been made from errors? Can you describe one?
Alex73 [517]

Answer:

the first popsicle was made because a kid accidentally left his fruit punch out in his front yard during winter

Explanation:

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3 years ago
A 9.13e+3 kg railroad car is rolling at 3.15 m/s when a 4.20e+3 kg load of gravel is suddenly dropped in from directly above. Wh
Feliz [49]

Answer:

6.85 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, since there are no external forces acting, the total momentum before and after must be conserved. So we can write:

m_1 v_1 = m_2 v_2

where

m_1 = 9.13\cdot 10^3 kg is the initial mass of the car

v_1 = 3.15 m/s is the initial speed of the car

m_2 = 9.13\cdot 10^3 kg - 4.20\cdot 10^3 kg=4.93\cdot 10^3 kg is the mass of the car after the load of gravel is dropped

v2 is the final speed of the car

Solving for v2, we find

v_2 = \frac{m_1 v_1}{m_2}=\frac{(9.13\cdot 10^3)(3.15 m/s)}{4.93\cdot 10^3}=6.85 m/s

6 0
3 years ago
a person takes a trip, driving with a constant speed of 89.5 km/h except for a 22.0 min rest stop. if the person’s average speed
fredd [130]

TheThe  distance they have covered for trip is 219Km.



<h3>What do you mean by uniformly accelereted motion?</h3>

When an object is traveling in a straight line with an increase in velocity at equal intervals of time.

The total time for the trip is

T→  t1+ 22 min = t1+ 0.367 h ,where t1 is the time spent traveling at

V1= 89.5 km/ h .

the distance traveled is ∆x = V1t1=Vavg T

after applying value and calculating it we get

 t1= 2.44h for a total time of

t total we get 2.81 h .

∆x =V1T1 = VavgTtotal

 ∆x = 77×2.81= 219 Km

to learn more about Uniform accelereted motion click here brainly.com/question/12920060

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3 0
2 years ago
A sail boat moves north for a distance of 10 km when blown by a wind 30° east of south with the force of 5.00×10^4 N. how much w
erma4kov [3.2K]
<h3><u>Answer;</u></h3>

a) 5.00 x 10^8 J

<h3><u>Explanation;</u></h3>

The work done to move the sailboat is calculated through the equation;

W = F x d

where F is force and d is the distance.

Substituting the known values from the given above,

                             W = (5.00 x 10⁴ N)(10 km)(1000 m/ 1km)

                                 = 5.00 x 10⁸ J

Thus, the work done is <u>5.00 x 10⁸Joules</u>

7 0
3 years ago
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