Respuesta :
Answer:
Internal energy will be [tex]2.05\times 10^5Joule[/tex]
Explanation:
We have given that initial volume [tex]v_1=1.53m^3[/tex]
And final volume [tex]v_2=2.87m^3[/tex]
Pressure [tex]p=1.02\times 10^5Pa[/tex]
Now change in volume [tex]dv=2.87-1.53=1.34m^3[/tex]
We know that change in internal energy is given by
[tex]\Delta U=\frac{3}{2}nR\Delta T[/tex], here we know that
Work done [tex]W=Pdv=nR\Delta T[/tex]
So [tex]\Delta U=\frac{3}{2}PdV=\frac{3}{2}\times 1.02\times 10^5\times 1.34=2.05\times 10^5Joule[/tex]
The change in internal energy should be [tex]2.05 \times 10^5 Joule[/tex]
Calculation of the change in the internal energy:
Since the initial volume is 1.53 m^3 to a final volume of 2.87 m^3
So, the change in the volume should be
= Final volume - initial volume
= 2.87 - 1.53
= 1.34
Now
Change in the internal energy should be
[tex]= \frac{3}{2} \times 1.02 \times 10^5 \times 1.34\\\\= 2.05 \times 10^5 Joule[/tex]
Learn more about the volume here: https://brainly.com/question/16836327?referrer=searchResults
