5 L of an ideal gas at a pressure of 20 atm expands
isothermally into a vacuum until its total volume is 10
L. The work done in the expansion process is
20 L atm
50 L atm
100 L atm
Zero
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Answer:
Hello
We know that in the expansion of any Gas, work is done by the system
W=−P
ext
Δv=−P
ext
(V
f
−V
i
)
p
ext
= external pressure of the system =0
In this Problem Pressure of 10atm is given and it is the pressure of Gas itself but we required external pressure in the expression and the gas is expanding into vacuum which has no pressure hence the external pressure =0
V
f
=10L
V
i
=2L
∵workdonebythesystem
W=−0×(10−2)=0
Hence, No work is done and heat absorbed (q) is a consequence of gaining temperature, but the system is working at constant temperature (Isothermal Condition). q=−w so, there will be no heat absorbed
(i)
P
ext
=1atm
V
f
=10L
V
i
=2L
∵W=−1atm×(10−2)L=−8atmL
Also, during isothermal expansion q=−w=8atmL
(ii)
Now the work is done then the expression of work done is
W
rev
=∫
V
i
V
f
P
in
dv
P
in
is the pressure of the Gas and not external pressure
Since, the gas is ideal, so PV=nRT or
P
in
=
V
nRT
W=∫
V
i
V
f
nRT
V
dV
=−nRTIn
V
i
V
f
=−2.303nRTlog
V
i
V
f
so,
P
in
=10atm=
V
nRT
V
f
=10L
V
i
=2L
W=−2.303nRTlog
V
i
V
f
=−2.303×20×log
2
10
=−32.2atmL
Q=−W=32.2 atm.L
Explanation:
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