1 mole of an ideal gas A (Cv,m=3R) and 2 mole of an ideal gas B (Cv,m=3/2R) taken in a container and expanded reversible and adiabatically from 1 litre to 4litre starting from intial temperature of 320K Delta U for process is??
Answers
Answer:
Correct option is
D
−960R
Solution:- (D) −960R
As we know that,
C
v
=
2
f
R
⇒f=
R
2C
v
Given:- C
v,m
A
=3⇒f
A
=6
C
v,m
B
=
2
3
R⇒f
B
=3
n
A
=1 mole
n
B
=2 mole
By using-
f
av.
=
n
A
+n
B
f
A
n
A
+f
B
n
B
⇒f
av.
=
1+2
1×6+2×3
⇒f
av.
=4
γ
av.
=1+
f
av.
2
⇒γ
av.
=1+
4
2
=
2
3
Now as we know that, for an adiabatic process,
TV
γ
av.
−1
=constant
⇒
T
1
T
2
=(
V
2
V
1
)
2
3
−1
Given:-
V
1
=1L
V
2
=4L
T
1
=320K
T
2
=T(say)=?
∴
320
T
=(
4
1
)
2
1
⇒T=160K
Again, as we know that,
ΔU=n
total
C
v
ΔT
As n
total
=1+2=3
C
v
=
2
f
av.
R=2R
ΔT=160−320=−160K
∴ΔU=3×2R×(−160)
⇒ΔU=−960R
Hence ΔE of ΔU for the process is −960R.
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Answer:
-960R
Explanation:
The answer is 960R
