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A gas is allowed to expand, at constant temperature, from a volume of 1.0 L to 10.1 L against an external pressure of
0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w, and a
9
AU

Respuesta :

nuhanoushad

Answer: A

Explanation:

q=heat absorbed=+250J

W=−PΔV=−0.5atm×(10−1)

=−4.5L−atm

=−4.5×101J=−454.5J

Now q=ΔE−W

ΔE=q+W=250−454.5

=−204.5J

pstnonsonjoku

The change in internal energy (ΔU) is -211 J.

According to the first law of thermodynamics;

ΔU = q + w

ΔU = change in internal energy

q = heat absorbed/ emitted

w = work done on or by the system

We can see from the question that  250 J of heat was absorbed from the surroundings hence q =  250 J

w= PΔV = 0.50 atm(10.1 L -  1.0 L) = 4.55 atm L

But;

1 L atm = 101.325 J

4.55 atm L =  4.55 atm L ×  101.325 J/ 1 L atm

= 461 J

Now;

ΔU = 250 J - 461 J

ΔU = -211 J

Note that work done is negative because the gas expands and does work on the surrounding.

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