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:
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
