Answer : The change in internal energy of the system is 15.62 kJ
Explanation :
First we have to calculate the work done of the system.
[tex]w=P\Delta V[/tex]
where,
w = work done
P = external pressure = 1.07 atm
[tex]\Delta V[/tex] = change in volume = 11.49 L
Now put all the given values on the above formula, we get:
[tex]w=1.07atm\times 11.49L[/tex]
[tex]w=12.3L.atm=12.3\times 101.3=1245.99J=1.25kJ[/tex] (as per conversion)
Now we have to calculate the change in internal energy of the system.
As per first law of thermodynamic,
[tex]\Delta E=q+w[/tex]
where,
[tex]\Delta E[/tex] = internal energy of the system
q = heat added to the system = 14.37 kJ
w = work done of the system = 1.25 kJ
Now put all the given values in the above formula, we get:
[tex]\Delta E=14.37+1.25[/tex]
[tex]\Delta E=15.62kJ[/tex]
Hence, the change in internal energy of the system is 15.62 kJ
The change in internal energy of the system is 13.125 KJ.
From the question, we have the following information;
Heat added to the system = 14.37 kJ
Pressure of the system = 1.07 atm
Volume change of the gas = 11.49 L
From the first law of thermodynamics, we know that;
ΔU = q + w
ΔU = change in internal energy of the system
q = heat absorbed
w = work done
Since the gas expanded, the system does work so w is negative
But w = -(PΔV) = -( 1.07 atm × 11.49 L) = -12.29 L.atm
Also;
1 L• atm = 101.3 J
-12.29 L.atm = -12.29 L.atm × 101.3 J/ 1 L• atm
= -1245 J
Since heat is added to the system, q is positive, therefore;
ΔU = 14.37 KJ + (-1.245 KJ)
ΔU =13.125 KJ
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