Question

Question: Consider a sample containing 2.00 moles of a monatomic ideal gas that undergoes the following changes:

For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ΔE, and ΔH for each step and for the overall change from state A to state D

Short answer

All the values regarding q, w, ΔE, and ΔH for each step and for the overall change from state A to state D has been shown below:

	Step 1	Step 2	Step 3	Overall change
q	-12.7 kJ	7.6 kJ	101.3 kJ	96.2 kJ
w	5.1 kJ	0	-40.5 kJ	-35.4 kJ
ΔE	-7.6 kJ	7.6 kJ	60.8 kJ	60.8 kJ
ΔH	-12.7 kJ	12.7 kJ	101.3 kJ	101.3 kJ

Step-by-step solution

Introduction

For monoatomic gas we can write

$\begin{array}{rcl}{C}_v& =& \frac{3}{2}R\\ C_p& =& \frac{5}{2}R\end{array}$

Let the work done by the system be denoted by the symbol w. The pressure and volume are denoted by P and V respectively. The amount of heat transferred is q. The internal energy be ΔE and enthalpy be ΔH.

Calculation for step1 given in problem

For Step 1

Given

$\begin{array}{rcl}{V}_A& =& 10.0L\\ V_B& =& 5.0L\\ P_A& =& 10.0atm\\ P_B& =& 10.0atm\\ n& =& 2.00mol\end{array}$

At constant pressure the following can be obtained

$\begin{array}{rcl}∆H& =& q_p\\ & =& n{C}_p∆T\\ & =& n\left(\frac{5}{2}R\right)\left(\frac{∆\left(PV\right)}{nR}\right)as\left[PV=nRT\&C_p=\frac{5}{2}R\right]\\ & =& \frac{5}{2}∆\left(PV\right)\\ & & \end{array}$

$$\begin{gathered} ∆H=\frac{5}{2}P∆V \\ =\frac{5}{2}\times 10atm\times \left(5.0-10.0\right)L \\ =-125L.atm\times 101.3L^{-1}at{m}^{-1} \\ =-12.7kJ \end{gathered}$$

$\begin{array}{rcl}w& =& -P∆V\\ & =& \left(-10.0atm\right)\left(5.0-10.0\right)L\\ & =& 5.1kJ\end{array}$

$\begin{array}{rcl}∆E& =& q+w\\ & =& -12.7+5.1\\ & =& -7.6kJ\end{array}$

Calculation for step2 given in problem

For Step 2

Given

$\begin{array}{rcl}{V}_A& =& 5.0L\\ V_B& =& 5.0L\\ P_A& =& 10.0atm\\ P_B& =& 20.0atm\\ n& =& 2.00mol\end{array}$

At constant volume the following can be obtained

$\begin{array}{rcl}∆E& =& q_v\\ & =& n{C}_v∆T\\ & =& n\left(\frac{3}{2}R\right)\left(\frac{∆\left(PV\right)}{nR}\right)as\left[PV=nRT\&C_v=\frac{3}{2}R\right]\\ & =& \frac{3}{2}∆\left(PV\right)\end{array}$

$$\begin{gathered} ∆E=\frac{3}{2}V∆P \\ =\frac{3}{2}\times 5atm\times \left(20.0-10.0\right)L \\ =-75L.atm\times 101.3L^{-1}at{m}^{-1} \\ =-7.6kJ \end{gathered}$$

$\begin{array}{rcl}w& =& -P∆V\\ & =& \left(-10\right)atm\left(5.0-5.0\right)L\\ & =& 0\end{array}$

$\begin{array}{rcl}∆E& =& q+w\\ & =& -12.7+5.1\\ & =& -7.6kJ\\ & & \end{array}$

$\begin{array}{rcl}∆H& =& ∆E+∆\left(PV\right)\\ & =& 75L.atm+50L.atm\\ & =& 12.7kJ\end{array}$

Calculation for step3 given in problem

For Step 3

Given

$\begin{array}{rcl}{V}_A& =& 5.0L\\ V_B& =& 25.0L\\ P_A& =& 20.0atm\\ P_B& =& 20.0atm\\ n& =& 2.00mol\end{array}$

At constant pressure the following can be obtained

$\begin{array}{rcl}∆H& =& q_p\\ & =& n{C}_p∆T\\ & =& n\left(\frac{5}{2}R\right)\left(\frac{∆\left(PV\right)}{nR}\right)as\left[PV=nRT\&C_p=\frac{5}{2}R\right]\\ & =& \frac{5}{2}∆\left(PV\right)\end{array}$

$\begin{array}{rcl}∆H& =& \frac{5}{2}P∆V\\ & =& \frac{5}{2}\times 20atm\times \left(25.0-5.0\right)L\\ & =& 1000L.atm\times 101.3L^{-1}at{m}^{-1}\\ & =& 101.3kJ\end{array}$

$\begin{array}{rcl}w& =& -P∆V\\ & =& -\left(20atm\right)\left(25-5\right)L\\ & =& -400L.atm\\ & =& -40.5kJ\end{array}$

$\begin{array}{rcl}∆E& =& q+w\\ & =& \left(101.3-40.5\right)kJ\\ & =& 60.8kJ\end{array}$

Determine the overall change

Now we need to calculate the overall changes from A to D

	Step 1	Step 2	Step 3	Overall change
q	-12.7 kJ	7.6 kJ	101.3 kJ	96.2 kJ
w	5.1 kJ	0	-40.5 kJ	-35.4 kJ
ΔE	-7.6 kJ	7.6 kJ	60.8 kJ	60.8 kJ
ΔH	-12.7 kJ	12.7 kJ	101.3 kJ	101.3 kJ