Question

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

Answer 1

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`