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Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics5 Marks
Question
A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. Find heat exchanged by the engine, with the surroundings for each section of the cycle. $(C_v = (3/2) R)$

AB : constant volume
BC : constant pressur CD : adiabati DA : constant pressure
✓

Answer

  1. : For A → B, dV = 0
so,$\text{dW}=\int\text{P}.\text{dV}=\int\text{p}\times0=0$
$\text{dW}=0$

By $1^\text{st}$ law of thermodynamics
$\text{dQ}=\text{dU}+\text{dW}=\text{dU}+0$
$\therefore\ \text{dQ} =\text{dU}$
$\big(\text{dQ}=\text{n}\text{C}_\text{V}\text{dT}\big)$
So, $\text{dQ}-1\frac{3}{2}\text{R}\big(\text{T}_\text{B}-\text{T}_\text{A}\big).....(\text{i})$
$\text{dU}=\text{dQ}=\frac{3}{2}\big(\text{R}\text{T}_\text{}B-\text{R}\text{T}_\text{A}\big)=\frac{3}{2}\big(\text{P}_\text{B}\text{V}_\text{B}-\text{P}_\text{A}\text{}V_\text{A}\big)$
$\therefore$ Heat exchange [to system]
$\text{dQ}_1=\text{dU}=\frac{3}{2}\big(\text{P}_\text{s}\text{V}_\text{s}-\text{P}_\text{A}\text{V}_\text{A}\big)$
  1. For B to C, $\Delta\text{P}=0\ \text{n}=1$
$\text{dQ}=\text{dU}+\text{dW}=\text{C}_\text{V}\text{(dT)}+\text{P}_\text{S}\text{dV}$
$\text{dQ}_2=\frac{3}{2}\text{R}\big(\text{T}_\text{C}-\text{T}_\text{B}\big)+\text{P}_\text{B}\big(\text{V}_\text{C}-\text{V}_\text{B}\big)$
$=\frac{3}{2}\big(\text{T}_\text{C}\text{R}-\text{RT}_\text{B}\big)+\text{P}_\text{B}\text{V}_\text{C}-\text{P}_\text{B}\text{V}_\text{B}$
$=\frac{3}{2}[\text{P}_\text{C}\text{V}_\text{C}]-\frac{3}{2}[\text{P}_\text{B}\text{V}_\text{B}]-\text{P}_\text{B}\text{V}_\text{B}-\text{P}_\text{B}\text{V}_\text{C}$
$\text{V}_\text{A}=\text{V}_\text{B}\ \text{and}\ \text{P}_\text{B}=\text{P}_\text{C}$
$\therefore\text{dQ}_2=\frac{3}{2}\text{P}_\text{B}\text{V}_\text{C}-\frac{3}{2}\text{P}_\text{B}\text{V}_\text{A}-\text{P}_\text{B}\text{V}_\text{A}+\text{P}_\text{B}\text{V}_\text{C}$
$=\frac{5}{2}\text{P}_\text{B}\text{V}_\text{C}-\frac{5}{2}\text{P}_\text{B}\text{V}_\text{A}$
$\text{dQ}_2=\frac{5}{2}\text{P}_\text{B}[\text{V}_\text{C}-\text{V}_\text{A}]$
  1. For diagram C → B, adiabatic change
$\text{dQ}_3=0$ (No exchange of heat )
  1. For diagram D → A, $\Delta\text{P}=0$ Compression of gas from volume $V_D$ to $V_A$ pressure hence heat exchange similar to part (b) i.e. Heat exchange$\text{dQ}_3=\frac{5}{2}\text{P}_\text{A}\big(\text{V}_\text{A}-\text{V}_\text{D}\big)$

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