Question

There is an expansion of 0.25 mº in a
certain volume of a gas at constant
pressure of 10' N/m2. Calculate the
work done against the pressure is
(A) 100 J
(B) 150 J
(C) 250 J
(D) 300 J​

Answer 1

Answer:

First change m^3 to lit 1m^3=10^3lit ; 0.25m^3 = 250lit W=-P[V] (-ve due to expansion) Now change pressure from pascal to atm 10^3pa=0.0098atm W= -0.0098*250 atmlit W= -2.25atmlit (1atm lit =101.33J) W= -2.25*101.33 W= 250J

I HOPE IT MAY HELPS YOU

Answer 2

Answer:

C) 250 J

ANSWER;

P=10

7

N/m

2

Δv=0.25

w=ρΔv

=10

3

×

100

25

=250J

∴ Option C is correct

I HOPE YOU ARE UNDERSTAND

PL MARK BRAINLIEST

THANKS