Consider a mixture of air and gasoline vapor in a cylinder with a piston and an original volume of 42.5 cm³. The combustion of this mixture releases 925 J of energy. What volume will the gas expand to against a constant external pressure of 1.05 atm if all the energy is converted into work pushing on the piston?

volume of the gas expands to 8.74 Litres

Explanation:

The amount of work done by the gas is given by;

W = P•ΔV

Where;

W = work done by gas

P = Pressure

ΔV = change in volume i.e V_f - V_i

Where V_f is final volume and V_i is initial volume

Thus;

W = P(V_f - V_i)

We are given;

W = 927J

V_i = 42.5 cm³ = 42.5 x 10^(-6) m³

P = 1.05 atm = 106391.25 N/m²

Thus;

925 = 106391.25(V_f - (42.5 x 10^(-6))

925/106391.25 = V_f - (42.5 x 10^(-6))

0.00869432402 = V_f - (42.5 x 10^(-6))

V_f = 0.00869432402 + (42.5 x 10^(-6)))

V_f = 0.00874 m³

Final volume = 0.00874 m³

Converting to Litres, gives;

V_f = 0.00874 x 1000 = 8.74 Litres

neither accurate nor precise.

inverse v=i*r

nice nice