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Energy and Enthalpy Changes, Heat and Work -- Monatomic Ideal Gas dynamically generated plot 2.00-mol of a monatomic ideal gas goes from State A to State D via the path A→B→C→D: State A PA=11.0atm, VA=12.50L State B PB=11.0atm, VB=7.00L State C PC=25.0atm, VC=7.00L State D PD=25.0atm, VD=20.50L Assume that the external pressure is constant during each step and equals the final pressure of the gas for that step. Calculate q for this process. q = 1pts Tries 0/6 Calculate w for this process. w = -902 J Use w = -PΔV and sum over the individual steps. Check that you have the correct energy units and be sure you have the sign right (w is defined as the work done ON THE SYSTEM). 1pts Incorrect. Tries 4/6 Previous Tries Calculate ΔE for this process ΔE = 1pts Tries 0/6 Calculate ΔH for this process. ΔH = 1pts

Answer :

Onyebuchinnajigo

(a) The heat generated in the process is 28 kJ.

(b) The work done in the process is determined as -28 kJ.

(c) The change in the internal energy is 0.

Heat of the isothermal compression

The heat generated in the process is negative done in the process.

W = -PΔV

W = -P(V₂ - V₁)

From A to B

W = -P(VB - VA)

W = -11(7 - 12.5)

W = 60.5 L.atm = 60.5 x 101.325 J/L.atm = 6,130.16 J

From C to D

W = -25(20.5 - 7)

W = -337.5 L.atm = -34,197.18 J

Total work , w = -34,197.18 J +  6,130.16 J = -28 kJ

q = - w

q = 28 kJ

Change in internal energy

ΔE = q + w

ΔE = 28 kJ - 28 kJ = 0

Learn more about change in internal energy here: https://brainly.com/question/17136958

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