Answer:
ΔU= 1.3922 KJ
Explanation:
Given that
Q= 2 KJ
Note- 1 .Heat added to the system taken as positive and heat leaving from the system is taken negative.
2. Work done on the system taken as negative and work done by the system taken as positive.
P =2 atm
V₁= 2 L
V₂=5 L
Work done by the gas W
W= P ( V₂- V₁)
W= 2 ( 5 - 2)
W= 6 atm.L
1 L·atm = 0.1013 kJ
W= 0.6078 KJ
From first law of thermodynamics
Q= W + ΔU
ΔU=Change in internal energy
2 = 0.6078 + ΔU
ΔU= 1.3922 KJ
The change in the internal energy of the system is 2.608 kJ.
The given parameters;
The work done on the system is calculated as;
W = PΔV
W = 2(5 - 2)
W = 6 Latm
1 atm = 0.1013 kJ
6 Latm = 0.608 kJ
The change in the internal energy of the system is calculated by applying first law of thermodynamics;
ΔU = Q + W
ΔU = 2 kJ + 0.608 kJ
ΔU = 2.608 kJ
Thus, the change in the internal energy of the system is 2.608 kJ.
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