Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e.
$$\eqalign{ & \eta = \frac{{{\text{Work done}}}}{{{\text{Heat supplied}}}} = \frac{W}{{{Q_1}}} = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}} \cr & = 1 - \frac{{{Q_2}}}{{{Q_1}}} = 1 - \frac{{{T_2}}}{{{T_1}}} \cr} $$
Here, $${{T_1}}$$ is the temperature of source, $${{T_2}}$$ is the temperature of sink, $${{Q_1}}$$ is heat absorbed and $${{Q_2}}$$ heat rejected
As given, $$\eta = 40\% = \frac{{40}}{{100}} = 0.4\,\,{\text{and}}\,\,{T_2} = 300\;K$$
So $$0.4 = 1 - \frac{{300}}{{{T_1}}} \Rightarrow {T_1} = \frac{{300}}{{1 - 0.4}} = \frac{{300}}{{0.6}} = 500\,K$$
Let temperature of the source be increased by $$x K,$$  then efficiency becomes
$$\eqalign{ & \eta ' = 40\% + 50\% {\text{ of }}\eta \cr & = \frac{{40}}{{100}} + \frac{{50}}{{100}} \times 0.4 \cr & = 0.4 + 0.5 \times 0.4 = 0.6 \cr} $$
Hence, $$0.6 = 1 - \frac{{300}}{{500 + x}}$$
$$\eqalign{ & \Rightarrow \frac{{300}}{{500 + x}} = 0.4 \cr & \Rightarrow 500 + x = \frac{{300}}{{0.4}} = 750 \cr & \therefore x = 750 - 500 = 250\;K \cr} $$
NOTE
All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot engine (as it is ideal).

Number of degree of freedom of a dynamical system is obtained by subtracting the number of independent relations from the total number of coordinates required to specify the positions of constituent particles of the system.
If $$A =$$  number of particles in the system,
$$R =$$  number of independent relations among the particles,
$$N =$$  number of degree of freedom of the system, then
$$N = 3A - R$$
Each monoatomic, diatomic and triatomic gas has three translatory degree of freedom.

As $$\Delta U = 0$$   in a cyclic process,
$$\eqalign{ & \Delta Q = \Delta W = {\text{area}}\,{\text{of}}\,{\text{circle}} = \pi {r^2} \cr & {\text{or}}\,\Delta W = {10^2}\pi J \cr} $$

$$\eqalign{ & {Q_1} = {T_0}{S_0} + \frac{1}{2}\,{T_0}{S_0} \cr & = \frac{3}{2}\,{T_0}{S_0} \cr & {Q_2} = {T_0}\left( {2\,{S_0} - {S_0}} \right) \cr & = {T_0}{S_0} \cr & {\text{and }}{Q_3} = 0 \cr} $$

$$\eqalign{ & \eta = 1 - \frac{{{Q_2}}}{{{Q_1}}} \cr & = 1 - \frac{{{T_0}{S_0}}}{{\frac{3}{2}\,{T_0}{S_0}}} \cr & = \frac{1}{3} \cr} $$

In isochoric process, the curve is parallel to $$y$$-axis because volume is constant. Isobaric is parallel to $$x$$-axis because pressure is constant. Along the curve, it will be isothermal because temperature is constant.
So, P → c ⇒ Q → a ⇒ R → d ⇒ S → b

$$\eqalign{ & ds = n{C_v}dT + PdV = 0 \cr & nR\frac{{dT}}{{dV}} + \left( {{p_0} - \alpha V} \right) = 0 \cr & pV = nRT \cr & {p_0}V - \alpha {V^2} = nRT \cr & {p_0} - 2\alpha V = nR\frac{{dT}}{{dV}} \cr & - \left( {{p_0} - \alpha V} \right)\left( {\gamma - 1} \right) = {p_0} - 2\alpha V \cr & - {p_0}\left( {\gamma - 1} \right) + \alpha \left( {\gamma - 1} \right)V = {p_0} - 2\alpha V \cr & {p_0}V = \alpha V\left( {\gamma + 1} \right) \cr & V = \frac{{{p_0}\gamma }}{{\alpha (\gamma + 1)}} \cr & V = \frac{{{p_0} \times \frac{5}{3}}}{{\alpha \left( {\frac{5}{3} + 1} \right)}} = \frac{{5{p_0}}}{{8\alpha }} \cr} $$

In reversible cyclic Process
$$\Delta U = 0$$

For adiabatic process $$P{V^\gamma } = {\text{constant}}$$
Also for monoatomic gas $$\gamma = \frac{{{C_P}}}{{{C_V}}} = 1.67$$
for diatomic gas $$\gamma = 1.4$$

Since, $${\gamma _{{\text{diatomic}}}} < {\gamma _{{\text{mono atomic}}}}$$
$$\therefore \,\,{P_{{\text{diatomic}}}} > {P_{{\text{mono atomic}}}}$$
⇒ Graph 1 is for diatomic and graph 2 is for mono atomic.

The molecule of a triatomic gas has a tendency of rotating about any of three coordinate axes. So, it has 6 degrees of freedom, 3 translational and 3 rotational. At high enough temperature a triatomic molecule has 2 vibrational degree of freedom. But as temperature requirement is not given, so we answer simply by assuming triatomic gas molecule at room temperature.
Thus, $$f = 6$$

(3 translational +3 rotational) at room temperature.

$$\eqalign{ & Q = n{C_P} \times 30 = n \times \frac{{7R}}{2} \times 30 \cr & {\text{and }}Q = n \times \frac{{5R}}{2} \times \Delta T \cr & \therefore n \times \frac{{7R}}{2} \times 30 = n \times \frac{{5R}}{2} \times \Delta T\,\,{\text{or}}\,\,\Delta T = 42\,K. \cr} $$