Question

A gas in an engine cylinder expands from a volume of 10.0 L to 15.0 L against an external pressure of 1 atm and the system absorbs 300 J of heat in the process. Determine the work done by the system and the change in the system's internal energy, both in joules. Use this conversion scale to calculate the work done in joules: 1 L * atm

Answer 1

Answer: The work done by the system is -506.5 Joules and change in the system's internal energy is -206.5 Joules

Step-by-step explanation:

According to first law of thermodynamics:

`\Delta E=q+w`

`\Delta E`=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=
`-P\Delta V` {Work is done by the system as the final volume is greater than initial volume and is negative}

w =
`-1atm* (15.0-10.0)L=-5.00Latm=-506.5Joules` {1Latm=101.3J}

q = +300J {Heat absorbed by the system is positive}

`\Delta E=+300+(-506.5)=-206.5J`

Thus the work done by the system is -506.5 Joules and change in the system's internal energy is -206.5 Joules