Question

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 45.0 atm and releases 71.4 kJ of heat. Before the reaction, the volume of the system was 7.60 L . After the reaction, the volume of the system was 2.20 L .

Calculate the total internal energy change, ΔE, in kilojoules.

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An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2)  A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 4.40 to 2.20 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.20 to 1.76 L .

In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 4.40 to 1.76 L in one step.

If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

Answer 1

ΔH     = ΔE + P Δv

ΔH = -71.4Kj

ΔE    = ΔH-P ΔV

        = -71.4 -45*101.325Kpa*(3*10^-3 m³ -7.6*10^-3 m³ )

      = -71.4 -45*101.325*(0.003-0.0076) = -50.42KJ

q1 = PV

       = 2*(4.4-2.2) = 4.4L-atm

q2 = PV

      = 2.5*(2.2-1.76) = 1.1L-atm

Total q = q1+q2

                = 4.4+1.1 = 5.5L-atm = 5.5*101.325j = 557.3J