[1] **Ideal Gas EoS**: #\quad PV=nRT;# (always true)

[2] **Monoatomic ideal gases**: If #U# is the internal energy, #C_p# and #C_v# are molar heat capacities at constant volume and constant pressure, respectively,

#U=3/2 nRT#

#C_p = 5/2R;\qquad C_v=3/2R;\qquad \gamma=C_p/C_v=5/3;#

[3] **Adiabatic Process**:

**EoS**: #\quad PV^\gamma = const.# (true only for adiabatic process)

**Work Done**: #W_{i\rightarrowf}=(p_fV_f-p_iV_i)/(\gamma-1)#

**Problem**

The three intermediate points #a#, #b# and #c# are characterized by the state variables #(p_a, V_a); \quad(p_b, V_b)# and #(p_c, V_c)#.

Observe that: #\quad V_a=V_b# and #\quadp_a=p_c#

**Step 1** (Adiabatic Expansion): During adiabatic process the system does not exchange heat with the surroundings.

#Q_{bc}=0;#

#p_bV_b^\gamma = p_cV_c^\gamma; \qquad \rightarrow p_c/p_b = (V_b/V_c)^\gamma#

Remember that #\quad p_a=p_c# and #V_c=8V_b=8V_a#,

#p_c/p_b=1/8^\gamma=p_a/p_b#......(**Eq1**)

Next calculate the work done -

#W_{bc} = (p_cV_c-p_bV_b)/(\gamma-1)=(p_bV_b)/(\gamma-1)[(p_cV_c)/(p_bV_b)-1]#

#W_{bc}=-(p_bV_b)/(\gamma-1)[1-1/8^(\gamma-1)]#......(**Eq2**)

Work done is negative, as is expected for expansion. Work is being done by the gas.

**Step 2** (Isobaric Compression): For isobaric processes the ratio of volume to temperature is a constant (Charles's law).

#V_c/T_c = V_a/T_a; \qquad \rightarrow T_c/T_a = V_c/V_a=8#

Change in internal energy -

#\Delta U_{ca}=3/2nR\DeltaT_{ca}=3/2nR(T_a-T_c)=3/2nRT_a(1-T_c/T_a)#

#\Delta U_{ca}=3/2nRT_a(1-8)=-21/2(nRT_a)#

Using the ideal gas equation of state, #\quad p_aV_a=nRT_a# and recognizing that #V_a=V_b#,

#\DeltaU_{ca}=-21/2p_aV_a=-21/2p_aV_b#

Using **Eq1** : #\quadp_a=p_b/8^\gamma#

#\DeltaU_{ca} = -21/2(p_b/8^\gamma)V_b=-21/2(p_bV_b)/8^\gamma#......(**Eq3**)

Work done during an isobaric process is-

#W_{ca}=-p_a\DeltaV_{ca}=-p_a(V_a-V_c)=p_a(V_c-V_a)#

#W_{ca}=p_aV_a(V_c/V_a-1)=p_aV_a(8-1)=7p_aV_a#

Using **Eq1**: #\quad p_a=p_b/8^\gamma# and the fact that #V_a=V_b#,

#W_{ca} = 7(p_b/8^\gamma)V_b=7(p_bV_b)/8^\gamma#......(**Eq4**)

Work done is positive, as it is expected for compression. Work is being done on the gas.

Next apply the first law of thermodynamics to calculate the heat,

#\Delta U_{ca}=Q_{ca}+W_{ca}; \qquad Q_{ca}=\DeltaU_{ca}-W_{ca}#

#Q_{ca} = -(21/2 +7)(p_bV_b)/8^\gamma=-35/(2\times8^\gamma)p_bV_b# ......(**Eq 5**)

Negative value for #Q_{ca}# indicates that **heat leaves the system**.

**Step 3** (Isochoric Heating): For an isochoric process volume is a constant and the ratio of pressure to temperature is a constant. Since the volume of the gas does not change, the work done during an isochoric process is zero.

#W_{ab}=0;#

#p_a/T_a = p_b/T_b; \qquad \rightarrow T_a/T_b = p_a/p_b=1/8^\gamma;#

The temperature the gases raises from #T_a# to #T_b#. So heat must have entered the system. Use the temperature difference and molar heat capacity for constant volume process to find this heat

#Q_{ab} = nC_v\Delta_{ab}=nC_v(T_b-T_a)=nC_vT_b(1-T_a/T_b)#

#Q_{ab}=n(3/2R)T_b(1-1/8^\gamma)=3/2(nRT_b)(1-1/8^\gamma)#

Using the ideal gas EoS : #\quadp_bV_b=nRT_b#,

#Q_{ab} = 3/2(1-1/8^\gamma)p_bV_b# ...... (**Eq 6**)

As expected, #Q_{ab}# is positive, indicating that **heat is added to the system**.

**Numerical Calculation**:

#p_b = 12\quadatm = (12 \quadcancel(atm))\times(1.01325\times10^5\quad (Pa)/(cancel{atm}))#

#p_b = 1.21\times10^6\quad N/m^2; \qquad V_b = 3.60\times10^{-3}\quadm^3;#

#p_bV_b = 4.35\times10^3\quadN.m = 4.35 \quadkJ#.

#\gamma = 5/3#

(a) **Heat added to the gas**:

#Q_{ab}=3/2(1-1/8^\gamma)p_bV_b = +6.32\quad kJ;#

(b) **Heat leaving the gas**:

#Q_{ca}=-35/(2\times8^\gamma)p_bV_b = -2.38\quad kJ;#

(c) **Net Work Done**:

#W_{bc}=-(p_bV_b)/(\gamma-1)[1-1/8^{\gamma-1}]=-4.89\quad kJ;#

#W_{ca}=+0.95\quad kJ;#

#W_{"net"}=W_{bc}+W_{ca}=-3.94\quad kJ#

Negative sign is because the work is done by the gas. Negative sign is as a result of sign convention, according to which the work done on the gas is positive and the work done by the gas is negative.

(d) **Efficiency of the cycle**:

#\eta\equiv |W|/Q_{"in"}=(|Q_{"in"}|-|Q_{"out"}|)/Q_{"in"}=1-|Q_{"out"}|/Q_{"in"}=1-|Q_{ca}|/Q_{ab}#

#\eta=62.3%#