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When 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston, under an external pressure of 2.00 atm, the nitrogen gas expands from 2.00 L to 5.00 L.

What is the change in internal energy of this system?

(Note: 1 L·atm = 0.1013 kJ)

Answer :

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

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Final answer:

The change in internal energy of the system, when 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder and the gas expands from 2.00 to 5.00 L under an external pressure of 2.00 atm, is calculated to be 1.392 kJ.

Explanation:

The change in internal energy of the system can be calculated using the formula derived from the first law of thermodynamics, which states that the change in internal energy (ΔE) of a system is equal to the heat transferred into the system (Q) minus the work done by the system (W). In this case, ΔE = Q - W. Heat transferred into the system (Q) is given as 2.00 kJ. The work done by the system can be calculated using the formula W = PΔV, where P is the external pressure and ΔV is the change in volume. Given that P = 2.00 atm and ΔV = 3.00 L (5.00 L - 2.00 L), and 1 L·atm = 0.1013 kJ, we find that W = 2.00 atm * 3.00 L * 0.1013 kJ/L·atm = 0.608 kJ. So, the change in internal energy is ΔE = Q - W = 2.00 kJ - 0.608 kJ = 1.392 kJ.

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