A helium gas balloon is expanded to 78.0 L, while the pressure is held constant at 0.37 atm. If the work done on the gas mixture was 28.2 J, what was the initial volume (in liters) of the helium gas balloon. (1 L. atm?

The initial volume of the helium gas balloon was 1.78 L

Explanation:

Step 1: Data given

Volume of the balloon is expanded to 78.0 L

The pressure is held constant at 0.37 atm

If the work done on the gas mixture was 28.2 J

Step 2: Calculate the initial volume

W = pΔV

⇒W = the work done on the gas = 28.2 J

⇒p = the pressure = 0.37 atm

⇒ΔV = the change in volume = V2 - V1 = 78.0 L - V1

W = 0.37 * ( 78.0 - V1)

28.2 J = 0.37 * ( 78.0 - V1)

28.2 J = 28.86 - 0.37V1

-0.66 = -0.37V1

V1 = 1.78 L

The initial volume of the helium gas balloon was 1.78 L

law of conservation of matter

b.final