Question

4.An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the working substance and operates from a most compressed stage of 100 Nm-2 and 327 K. It expands isothermally to a pressure of 90 Nm-2 and then adiabatically to a most expanded stage of 27 K. Calculate the ΔU, q, and w for each step. Calculate the net work done and the efficiency of the cycle [Cv,m for the gas] is 25 J/k/mol.

Answer 1

Answer:

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Answer 2

Given :

$T_1=T_2=327K$

$T_3=T_4=27K$

$P_1= 100 \frac{N}{m^2}$

$P_2= 90 \frac{N}{m^2}$

$C_v= 25 J/k/mol$

Amount of gas, n= 1 mol

Carnot Cycle

To Find:

(a) Heat Transfer ($\delta q$), Work done ($\delta w$) and Internal Energy Change ($\triangle U$) for each step ie for each process of the Carnot cycle.

(b) Net work done, W

(c) Efficiency, $\eta$

Explanation:

Carnot Cycle consists of Two Isothermal processes ($PV=c$)and two adiabatic ($PV^\gamma = c$) processes. Say, the Carnot cycle is 1-2-3-4, so the processes are 1-2 (Isothermal expansion), 2-3 (adiabatic expansion), 3-4 (isothermal compression) and 4-1 (adiabatic compression).

(a) Process 1-2 Isothermal expansion

Work done, $\delta w_1= n\times R\times T_1\times ln\frac{P_1}{P_2} J$,

$\\\delta w_1= 1\times 8.314\times 327 \times ln\frac { 100}{90} J\\\delta w_1= 286.44 J$

Also. we know by First Law of thermodynamics that $\delta q = \triangle U + \delta w$

and for an isothermal process, $\triangle U =0$,

$\therefore$ $\delta q_1= \delta w_1= 286.44 J$ ($\delta q_1$ is +ve because heat is supplied to the system)

(b)  Process 2-3 Adiabatic expansion

Work done, $\delta w_2= \frac {n\times R\times (T_2-T_3) }{\gamma-1}$

$\delta w_2=\frac {1\times 8.314\times (327-27)}{1.4-1}\\\delta w_2= 6235.5 J$

For an adiabatic process, $\delta q=0$

$\therefore, \triangle U_2=-\delta w_2= - 6235.5 J$($\Rightarrow$ expansion is being done at the cost of internal energy)

(c) Process 3-4 Isothermal Compression

Work done, $\delta w_3= n\times R\times T_3\times ln\frac{P_3}{P_4} J$

Now, we know that for an adiabatic process 2-3 ,

$\frac {T_2}{T_3}= (\frac {P_2}{P_3})^ \frac {\gamma-1}{\gamma}\\\frac {327}{27}= (\frac{90}{P_3})^\frac{1.4-1}{1.4}\\\Rightarrow P_3= 0.14 \frac {N}{m^2}$

and or an adiabatic process 4-1

$\frac {T_4}{T_1}= (\frac {P_4}{P_1})^ \frac {\gamma-1}{\gamma}\\\frac {27}{327}= (\frac{P_4}{100})^\frac{1.4-1}{1.4}\\\Rightarrow P_4= 0.16 \frac {N}{m^2}$

$\therefore \\\delta w_3= 1\times 8.314\times 27 \times ln\frac { 0.14}{0.16} J\\\delta w_3= -29.97J$(Here, - ve sign denotes work is one on the system)

and $\delta q_3= \delta w_3= -29.97 J$ as $\triangle U_3=0$ for isothermal process.

(c) Process 4-1 Adiabatic Compression

Work done, $\delta w_4= \frac {n\times R\times (T_4-T_1) }{\gamma-1}$

$\delta w_4=\frac {1\times 8.314\times (27-327)}{1.4-1}\\\delta w_2= -6235.5 J$

For an adiabatic process, $\delta q=0$

$\therefore, \triangle U_4=-\delta w_4= 6235.5 J$

$\therefore$ Net work done,

$W= \delta w_1 + \delta w_2 + \delta w_3 + \delta w_4\\W= 256.47 J$

Efficiency,

$\eta= 1-\frac {T_4}{T_1}\\ = 1- \frac {27}{327}\\ = 0.917\\ = 91.7 %$%