A sample of nitrogen gas expands in volume from 1.6l to 5.4l at constant temperature what is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?
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0
Answer:
Work done by the gas in isothermal expansion w = -p. DV
p = pressure, DV = final volume (V2) - initial volume (V1)
(a) When expand against vacuum then p= 0 atm
W = -0 atm×(5.4-1.6) L = 0 L.atm = 0 J
(b) Here, p = 3.7 atm
V2-V1 = ( 5.4-1.6) L = 3.8 L
1 L.atm = 101.325 J
w = - 3.7×3.8 L.atm
= -14.06 L.atm
= -14.06×101.325 J
= -1424.63 J
Answered by
0
Answer:
a=0j
b=-308.03 J
Explanation:
(a) When expand against vacuum then p= 0 atm
W = -0 atm×(5.4-1.6) L = 0 L.atm = 0 J
(b) Here, p = 0.80 atm
V2-V1 = ( 5.4-1.6) L = 3.8 L
1 L.atm = 101.325 J
w = - 0.80×3.8 L.atm
= -3.04 L.atm
= -3.04×101.325 J
= -308.03 J
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