# Cylinder-and-piston physics probem

1. Dec 7, 2005

### jakeowens

A cylinder-and-piston containing an ideal gas is placed in contact a thermal reservoir. The volume of the gas is very slowly changed from 55 liters to 8 liters as 50 J of work is done on it by an external agency. Determine the change in the internal energy of the gas

and the amount of heat flowing into or out of the system

For this one, I used the first law of thermodynamics. Delta U=Qinto system - W done by system. Since work is done to the system, its minus a negative so Delta U= Q+W. Where do i go from here though? i know the work done, but i dont know the heat flowing in (in because its being compressed) or the change in energy. Do i somehow use the ideal gas law? im just kind of lost on what i need to do next.

A gas in a thin plastic bag at atmospheric presure receives 9640 J of heat and, in the process, puffs up the bag, increasing its volume by 1.05 m3. By how much is the internal energy of the gas altered?

For this one i used the equation Work done by system= Pressure * delta V. So work done = 1.01x10^5 Pa *1.05m^3 = 106,050 J. Then using the first law of thermodynamics Delta U= 9630J-106050J = -96410J correct?

2. Dec 8, 2005

### Chi Meson

the second question you have correct.

FOr the first question, the trick is in the phrase "the volume is changed very slowly" from 55 to 8 liters. This is "code" for isothermal process. IF any thermal process is done very slowly, the heat will be able to flow into or out of a gas to keep the gas at the same temperature as its surroundings. If temperature does not change, then internal energy remains constant.

Remember, "very fast" indicates "adiabatic," "very slow" indicates "isothermal." I don't know why all texbooks don't come right out and say that in so many words.