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Consider a sample containing 5.00 moles of a monatomic ideal gas that is taken from state A to state B by the following two pathways: For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ∆E and ∆H for each step in kJ, and calculate overall values for each pathway. Explain how the overall values for the two pathways illustrate that ∆E and ∆H are state functions, whereas q and w are path functions. Hint: In a more rigorous study of thermochemistry, it can be shown that for an ideal gas: Δ E = nC v Δ T and Δ H = nC p Δ T where C v is the molar heat capacity at constant volume and C p is the molar heat capacity at constant pressure. In addition, for a monatomic ideal gas, C v = 3 2 R and C p = 5 2 R , where R = 8.3145 J/K ·mol.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 121MP
Textbook Problem
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Consider a sample containing 5.00 moles of a monatomic ideal gas that is taken from state A to state B by the following two pathways:

Chapter 16, Problem 121MP, Consider a sample containing 5.00 moles of a monatomic ideal gas that is taken from state A to state

For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ∆E and ∆H for each step in kJ, and calculate overall values for each pathway. Explain how the overall values for the two pathways illustrate that ∆E and ∆H are state functions, whereas q and w are path functions. Hint: In a more rigorous study of thermochemistry, it can be shown that for an ideal gas:

Δ E = nC v Δ T  and

Δ H = nC p Δ T

where Cv is the molar heat capacity at constant volume and Cp is the molar heat capacity at constant pressure. In addition, for a monatomic ideal gas, C v = 3 2 R and C p = 5 2 R , where R = 8.3145 J/K ·mol.

Interpretation Introduction

Interpretation: The values of q,w,ΔE and ΔH for each of the given steps in the two pathways in the given state change is to be calculated.

Concept Introduction: The internal energy change for the monoatomic ideal gas is calculated by using the formula,

ΔE=32nR(T2T1)

The enthalpy change of the monoatomic ideal gas is calculated by using the formula,

ΔH=52nR(T2T1)

To determine: The values of q,w,ΔE and ΔH for each of the given steps.

Explanation of Solution

Given

The pathway one for the change in state from A to B is,

PA=3.00atmVA=15.0L1PC=3.00atmVC=55.0LPC=3.00atmVC=55.0L2PB=6.00atmVB=20.0L

The pathway two for the change in state from A to B is,

PA=3.00atmVA=15.0L3PD=6.00atmVD=15.0LPD=6.00atmVD=15.0L4PB=6.00atmVB=20.0L

The number of moles of monoatomic gas in the sample is 5.00mol.

The temperature corresponding to each of the given states is calculated by using the ideal gas equation.

PV=nRTT=PVnR (1)

Where,

  • n is the number of moles of the gas.
  • P is the absolute pressure of the gas.
  • V is the volume of the gas.
  • T is the absolute temperature of the gas.
  • R is the gas constant.

Substitute the given values of P,V,n and R for state A in the equation (1).

TA=3.00atm×15.00L5.00mol×0.08206Latm/molK=109.68K

Substitute the given values of P,V,n and R for state B in the equation (1).

TB=6.00atm×20.00L5.00mol×0.08206Latm/molK=292.47K

Substitute the given values of P,V,n and R for state C in the equation (1).

TC=3.00atm×55.00L5.00mol×0.08206Latm/molK=402.14K

Substitute the given values of P,V,n and R for state D in the equation (1).

TD=6.00atm×15.00L5.00mol×0.08206Latm/molK=219.35K

The internal energy change for each of the given steps for the monoatomic gas is calculated by using the formula,

ΔE=32nR(T2T1) (2)

Where,

  • n is the number of moles of the gas.
  • R is the gas constant.
  • T2 is the final temperature.
  • T1 is the initial temperature.
  • ΔE is the internal energy change.

Substitute the values of n , R , T2 and T1 for the step 1 in the equation (2).

ΔE=32×5.00mol×8.3145J/molK×(402.14109.68)K=18237.4J(=18.237kJ)

Substitute the values of n , R , T2 and T1 for the step 2 in the equation (2).

ΔE=32×5.00mol×8

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