Question

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?

Answer 1

Answer:

77.248 L

Explanation:

From the question,

Work done on the gas mixture is given as,

W = PΔV.................. Equation 1

Where W = work done, P = pressure of the the gas, ΔV = Change in volume of the gas.

make ΔV the subject of the equation

ΔV = W/P..................... Equation 2

Given: W = 28.2 J, P = 0.37 atm = (0.37×101325) N/m² = 37490.25 N/m²

Substitute into equation 2

ΔV = 28.2/37490.25

ΔV = 0.000752 m³

ΔV = 0.752 L

But,

ΔV = V₂-V₁................. Equation 3

Where V₂ = Final volume of the helium gas, V₁ = Initial volume of the helium gas

make V₁ the subject of the equation

V₁ = V₂-ΔV................ Equation 4

Given: V₂ = 78 L.

Substitute into equation 4

V₁ = 78-0.752

V₁ = 77.248 L

Answer 2

Answer:

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