When 2.00 kj of energy is transferred as heat to nitrogen in a cylinder fitted with a piston with an external pressure of 2.00 atm, the nitrogen gas expands from 2.00 to 5.00 l. what is the change in internal energy of this system? (1 l·atm = 0.1013 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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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 right answer is 0.3 meters per second