Determine the change in internal energy, of the gas during this process

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 10K views
new^2^physics
Messages
13
Reaction score
0

Homework Statement



A gas sample heated in a cylinder used 650 kJ of heat. A piston that compresses the gas does 800 kJ of work. What is the change in internal energy, ΔE, of the gas during this process


Homework Equations



ΔE = q + w

The Attempt at a Solution



ΔE = q + w
ΔE = (650kJ) + (800 kJ)
ΔE = 1450 kJ

a result of the gas being heated and compressed, the energy of the gas increases by 1450 kJ. Is this at all correct? Or should it be -650 kJ? Thank you for your help.
 
Physics news on Phys.org
Hi new^2^physics, welcome to PF. Your answer looks fine.
 
new^2^physics said:
ΔE = q + w
ΔE = (650kJ) + (800 kJ)
ΔE = 1450 kJ

a result of the gas being heated and compressed, the energy of the gas increases by 1450 kJ. Is this at all correct? Or should it be -650 kJ?
To avoid problems, you have to use signs correctly. You should state the first law as:

[tex]\Delta Q = \Delta U + W[/tex]

where positive [itex]\Delta Q[/itex] is the heat flow INTO the gas, positive [itex]\Delta U[/itex] is the INCREASE in internal energy of the gas, and positive W is the work done BY the gas (so if work is done on the gas, W is negative).

If you make careful use of the signs (+ or -) you will avoid confusion.

In this case, you would state the first law as:

[tex]\Delta U = \Delta Q - W[/tex]

where [itex]\Delta Q = 800 J[/itex] and W = -650 J. (since work is done ON the gas).

AM