Answer to Question #270849 in Mechanics | Relativity for giel

Question

Gaseous nitrogen actuates a Carnot power cycle in which the respective volumes at the four corners of the cycle, starting at the beginning of the isothermal expansion, are 𝑉1 = 10.10 𝐿, 𝑉2 = 14.53 𝐿, 𝑉3 = 226.54 𝐿, 𝑉4 = 157.73 𝐿. The cycle receives 21.1 KJ of heat. Determine the work.

Answer 1

$V_1 = 10.1 \;L \\ V_2 = 14.53 \;L \\ V_3 = 226.54 \;L \\ V_4 = 157.73 \;L$

It is the case of reverce carnot cycle.

Heat resieved by cycle is

$Q_L = 21.1 \;kJ$

For process 2-3:

Heat interaction is

$Q_L = -W 21.1 \times 10^3 = nRT_2ln(\frac{V_3}{V_2}) \\ 21.1 \times 10^3 = 1 \times \frac{8314}{28} \times T_2 \times ln(\frac{226.54}{14.53}) \\ T_2 = 25.87 \;K$

Now, for process 3-4 is an adiabatic process

$\frac{T_H}{T_L} = (\frac{V_3}{V_4})^{Y-1} \\ Y=1.4$

for adiabatic

$\frac{T_H}{25.87} = (\frac{226.54}{157.73})^{0.4} \\ T_H = 29.901 \;K$

Now, heat transfer in process 1-4

$Q_{14} = -nRTln(\frac{V_1}{V_4}) \\ = -1 \times \frac{8314}{28} \times 29.901 \times ln(\frac{10.1}{157.73}) \\ = 24.401 \; kJ$

Net heat transfer $= Q_{12} +Q_{23} +Q_{34} +Q_{41}$

$= 0-21.1+0+24.401 \\ = 3.301 \;kJ$

In a cycle we know that

$\sum Q_{net} = \sum W_{net} \\ W_{net} = 3.301 \;kJ$