In a no flow reversible process for which p = (-3V + 15) × 105 N/m2, V changes from 1 m3 to 2 m3. The work done will be about
A. 100 × 105 joules
B. 1 × 105 joules
C. 10 × 105 joules
D. 10 × 105 kilo joules
Answer: Option C
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W=∫²₁(-3V+15)×10⁵dV
W=10.5×10⁵j
P=(-3v+15)x10^5 N/m^2
as V changes from 1-2=1
placing in equation
we get
p=12x10^5
so
work=PV
W=(12x10^5)(1)=12x!0^5 j
nearest vale is 10x10^5j
How?
can i get the solution for above problem