Answer:
The change in internal energy of the system is -17746.78 J
Explanation:
Given that,
Pressure [tex]P=4.63\times10^{4}\ Pa[/tex]
Remove heat [tex]\Delta U= -1.95\times10^{4}\ J[/tex]
Radius = 0.272 m
Distance d = 0.163 m
We need to calculate the internal energy
Using thermodynamics first equation
[tex]dU=Q-W[/tex]...(I)
Where, dU = internal energy
Q = heat
W = work done
Put the value of W in equation (I)
[tex]dU=Q-PdV[/tex]
Where, W = PdV
Put the value in the equation
[tex]dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))[/tex]
[tex]dU=-17746.78\ J[/tex]
Hence, The change in internal energy of the system is -17746.78 J
