Respuesta :
Answer: 292.95 J
Explanation:
change in internal energy= Heat transfer - work done
ΔU =Q -PΔV
Here, Q = 0 as there is no heat transfer.
P =2.00 atm = 2.00 × 101235 Pa = 202470 Pa
ΔV = final volume - initial volume = 0.8 V -V = -0.2 V
where V is the initial volume.
Volume of a spherical ball, [tex]V = \frac{4}{3}\pi r^3[/tex]
r = d/2 = 23.9 cm / 2 = 0.12 m
[tex]V = \frac{4}{3}\times 3.14 \times (0.12m)^3= 7.23\times10^{-3}m^3[/tex]
[tex]\DeltaU = -P\DeltaV = - 202470 Pa \times -0.2 \times 7.23\times10^{-3}m^3=292.95 J[/tex]
Hence, internal energy would change by 292.95 J.
The change in internal energy after the ball is compressed is [tex]\fbox{289.7\,{\text{J}}}[/tex].
Further explanation:
Change in energy is done on the cost of its internal energy which is given by the first law of thermodynamics.
Given:
The new volume of the ball is 0.8 times the original volume.
The pressure inside the ball is 2atm.
The diameter of the ball is 23.9cm.
Concept used:
Law of Conservation of Energy state that “Energy can neither be created nor be destroyed but only it can be transferred from one to another form and also called first law of thermodynamics.
The first law of thermodynamics state that "the amount of change in internal energy [tex]\Delta U[/tex] of one system is expressed as sum of heat [tex]Q[/tex] that transferring across its boundaries of the system and work done [tex]W[/tex] on system by surroundings":
[tex]\Delta U = Q + W[/tex] …… (1)
The expression for the work done is given as.
[tex]W = P\Delta V[/tex]
Here, [tex]\Delta V[/tex] is the change in volume and P is the pressure.
In this system there is no heat transfer i.e. [tex]Q = 0[/tex].
Substitute [tex]P\Delta V[/tex] for W and [tex]0[/tex] for Q in equation (1).
[tex]\fbox{\begin\\\Delta U = P\Delta V\end{minispace}}[/tex]
The final expression reduces as.
[tex]\Delta U = P\left( {{V_2} - {V_1}} \right)[/tex] …… (2)
Here, [tex]{V_1}[/tex] is the original volume, [tex]{V_2}[/tex] is the compressed volume and[tex]\Delta U[/tex] is the change in internal energy.
The expression for the volume of sphere is given as.
[tex]{V_1} = \dfrac{4}{3}\pi {r_1}^3[/tex] …… (3)
Here, [tex]{r_1}[/tex] is the radius of ball.
Substitute[tex]\left( {\frac{{23.9}}{2}} \right){\text{cm}}[/tex]for [tex]r_1[/tex] in equation (3).
[tex]\begin{aligned}{V_1}&=\frac{4}{3}\pi {\left( {\frac{{23.9\,{\text{cm}}}}{2}} \right)^3}\\&=\frac{4}{3}\pi {\left( {\frac{{23.9\,{\text{cm}}}}{2}\left( {\frac{{1\,{\text{m}}}}{{100\,{\text{cm}}}}} \right)} \right)^3} \\&=7.15\,{{\text{m}}^{\text{3}}}\\ \end{aligned}[/tex]
Substitute [tex]0.8{V_1}[/tex] for [tex]{V_2}[/tex] , [tex]2\,{\text{atm}}[/tex] for P and [tex]7.15\,{{\text{m}}^{\text{3}}}[/tex] for [tex]{V_1}[/tex] in equation (2).
[tex]\begin{aligned}\Delta U&=\left( {2\,{\text{atm}}} \right)\left( {0.8{V_1} - {V_1}} \right) \\&=\left({2\,{\text{atm}}\left( {\frac{{{\text{101325}}\,{\text{Pa}}}}{{1\,{\text{atm}}}}} \right)} \right)\left( {0.2{V_1}} \right)\\&=\left( {202650} \right)\left( {0.2\left( {7.15\,{{\text{m}}^{\text{3}}}} \right)} \right)\\&=289.7\,{\text{J}} \\ \end{aligned}[/tex]
Thus, the change in internal energy of the ball is [tex]\fbox{289.7\,{\text{J}}}[/tex].
Learn more:
1. Example of energies https://brainly.com/question/1062501.
2. Motion under friction https://brainly.com/question/7031524.
3. Average translational kinetic energy https://brainly.com/question/9078768.
Answer Details:
Grade: College
Subject: Physics
Chapter: Heat and Thermodynamics
Keywords:
Energy, heat, work, first law of thermodynamics, conservation of energy, volume, mass, change in volume, heat transfer, compression, system, surrounding, 289.7J, 290J, 7.15m^3.
