Question

A gas undergoes a cycle in a piston–cylinder assembly consisting of the following three processes: Process 1–2: Constant pressure, p = 1.4 bar, V1 = 0.028 m3, W12 = 15 kJ Process 2–3: Compression with pV = constant, U3 = U2 Process 3–1: Constant volume, U1 - U3 = -25 kJ There are no significant changes in kinetic or potential energy. (a) Calculate the net work for the cycle, in kJ. (b) Calculate the heat transfer for process 1–2, in kJ.

Answer 1

Answer:

(a) -5.082 kJ

(b) 40 kJ

Explanation:

(a)

The net work of the cycle is calculated as

$w_{cycle} = w_{12} + w_{23} + w_{31} (1)$

$w_{31} = 0$ because $V_3 = V_1$ and $W_{12} = 15 kJ$, so $w_{23}$ has to be computed.

$w_{23} = \int_{V_2}^{V_3} pdV = \int_{V_2}^{V_3} \frac{constant}{V} dV = p_2 V_2 \, ln \frac{V_3}{V_2} (2)$

From data

$V_3 = V_1 = 0.028 m^3$

$p_2 = p_1 = 1.4 bar = 1.4\times10^5 Pa$

We need to find $V_2$, to do so

$w_{12} = \int_{V_1}^{V_2} pdV = p_1 \, (V_2 - V_1)$

$V_2 = \frac{w_{12}}{p_1} + V_1$

$V_2 = \frac{15\times10^3 J}{1.4\times10^5 Pa} \times \frac{N \, m}{J} \times \frac{Pa}{N/m^2} + 0.028 m^3$

$V_2 = 0.107 m^3$

From equation 2

$w_{23} = p_1 V_2 \, ln \frac{V_1}{V_2}$

$w_{23} = 1.4\times10^5 N/m^2 \, 0.107 m^3 \, ln \frac{0.028 m^3}{0.107 m^3}$

$w_{23} = -20082 J = -20.082 kJ$

From equation 1

$w_{cycle} = w_{12} + w_{23} + w_{31}$

$w_{cycle} = 15 kJ -20.082 kJ$

$w_{cycle} = -5.082 kJ$

(b)

Taking into account that there are no significant changes in kinetic or potential energy, from the energy balance in 1-2 process we get

$U_2 - U_1 = Q_{12} - W_{12}$

With $U_2 = U_3$

 

$Q_{12} = (U_3 - U_1) + W_{12}$

$Q_{12} = 25 kJ+ 15 kJ$

$Q_{12} = 40 kJ$

Answer 2

Based on the calculations, the net work for this cycle is equal to -20.12 kJ.

How to calculate the net work for the cycle.

• Pressure = 1.4 bar.

• $V_1=0.0028\;m^3$

• $W_{12}=15\;kJ$

Since process 1-2 has a constant pressure, the work done is given by this formula:

$\Delta W=P\Delta V=P(V_2-V_1)\\\\15\times 10^3=1.4\times 10^5(V_2-0.028)\\\\15000=140000V_2-3920\\\\140000V_2=15000+0.0392\\\\V_2=\frac{15000.0392}{140000} \\\\V_2=0.1071\;m^3$

During process 2-3, we have:

$pV = constant\\\\U3 = U2$

During process 3-1, we have:

$U1 - U3 = -25 kJ\\\\V=k$

We would calculate the final temperature ($T_2$) by using the ideal gas equation:

$P_2V_2=RT_2\\\\T_2=\frac{P_2V_2}{R} \\\\T_2=\frac{1.4 \times 10^5 \times 0.1071}{8.314}\\\\T_2=1,803.46\;K$

For the work done:

$W_{23}=RT_2ln\frac{V_1}{V_2} \\\\W_{23}=8.314 \times 1803.46 \times ln\frac{0.028}{0.1071}\\\\W_{23}= 14994 \times ln(0.2614)\\\\W_{23}= 14994 \times ( -1.3417)\\\\W_{23}= -20117.50\\\\W_{23}= -20.12\;kJ$

Now, we can determine the net work for the cycle:

$\Delta W_{net}=-20.12 + 15\\\\\Delta W_{net}= -5.12\;kJ$

How to calculate the the heat transfer for process 1-2.

$\Delta Q=\Delta U + W\\\\\Delta Q=25+15\\\\\Delta Q=40\;kJ$

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