One mole of an ideal gas at temperature T was cooled
isochorically till the gas pressure fell from P to P/n. Then,
by an isobaric process, the gas was restored to the initial
temperature. The net amount of heat absorbed by the gas
in the process is
(a) nRT (b) RT/n
(c) RT (1 – n⁻¹) (d) RT (n – 1)
Answers
Answered by
1
Answer:
a is your answer
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Answered by
1
Answer:
The temperature remains unchanged therefore
U
f
=U
i
Also, ΔQ=ΔW.
In the first step which is isochoric, ΔW=0.
In second step, pressure =
n
P
. Volume is increased from V to nV.
∴W=
n
P
(nV−V)
= PV(
n
n−1
)
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