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When 4.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston at an external pressure of 3.00 atm, the nitrogen gas expands from 1.00 to 4.00 L against this constant pressure. What is U for the process?

1. -4.91 kJ
2. -0.912 kJ
3. +3.09 kJ
4. 0
5. +4.91 k

User NYCBilly
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1 Answer

4 votes

Answer:

3. +3.09 kJ

Step-by-step explanation:

The change in internal energy of the gas is given by the 1st law of thermodynamics:


\Delta U=Q-W

where

Q is the heat transferred to the gas

W is the work done by the gas

Here we have:


Q=+4.00 kJ is the amount of heat transferred to the nitrogen


p=3.00 atm = 3.03\cdot 10^5 Pa is the pressure of the gas


\Delta V=4.00 L-1.00 L=3.00 L = 0.003 m^3 is the change in volume of the gas

So the work done by the gas is


W=p\Delta V=(3.03\cdot 10^5 Pa)(0.003 m^3)=909 J = 0.91 kJ

So, the change in internal energy of the gas is


\Delta U=4.00 kJ-0.91 kJ=+3.09 kJ

User Utdemir
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