work done by air when it expands from 50 lts. to 150 lts. at a constant pressure of 2 atmosphere?

Question

Please write the solution in detail.

Answer 1

Hmmm
You want a detailed answer then here you go

Think of a piston that has an area A to which a force F is applied as a result of a given pressure (in this case 2 atm).

Then work done W=Fd d being the distance the piston had to travel.

d=V/A

V=Differential Volume = 150-50=100 liters

F=PA where P - pressure and A = area

Then W=PA V/A= PV

1atm = 101 325 N/m^2
1ltr=10e-3m^3
then work done is for 100 liters
W=P V=2(101325 )(100)(10e-3m^3)
W=20266 joules

Answer 2

Use the formula: W=pV

W is the work done, p is the pressure and V is the change in volume.

First, calculate the change in volume: 150 - 50 = 100 l

Now, put it into the formula: 2 * 100 = 200 Joules

Answer 3

P1=x
P2=2 atmospheres
V1=50 lts.
V2=150-50=100 atmospheres.
By Boyles law,
P1V1=P2V2
x.50=2.100
4 atmospheres.

Answer 4

the work done by air, dW= P*dV

here,
P= 2 atm= 2*101325 Pa =202650 Pa
dV= (150-50) lts.= 100 lts.= 0.1 m^3

thus,
dW= 202650*0.1=20.265 kJ

Answer 5

Same as avik above. But i do not know why the symbol dw is used instead of w?. dw notation implies difference in work done.