Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & {T_1} = 273 + 27 = 300\,K \cr & {T_2} = 273 + 927 = 1200\,K \cr} $$
For adiabatic process,
$$\eqalign{ & {P^{1 - \gamma }}{T^\gamma } = {\text{constant}} \Rightarrow {P_1}^{1 - \gamma }{T_1}^\gamma = {P_1}^{1 - \gamma }{T_2}^\gamma \cr & \Rightarrow {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{1 - \gamma }} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^\gamma } \Rightarrow {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{1 - \gamma }} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^\gamma } \cr & {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{1 - 1.4}} = {\left( {\frac{{1200}}{{300}}} \right)^{1.4}} \Rightarrow {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{ - 0.4}} = {\left( 4 \right)^{1.4}} \cr & {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{0.4}} = {4^{1.4}}\,{\text{or,}}\,{P_2} = {P_1}{4^{\left( {\frac{{1.4}}{{0.4}}} \right)}} = {P_1}{4^{\left( {\frac{7}{2}} \right)}} \cr & = {P_1}\left( {{2^7}} \right) = 2 \times 128 = 256\,atm \cr} $$

In the first process $$W$$ is $$+ ve$$  as $$\Delta V$$ is positive, in the second process $$W$$ is $$- ve$$  as $$\Delta V$$ is $$- ve$$  and area under the curve of second process is more
∴ Net Work $$< 0$$  and also $${P_3} > {P_1}.$$

$$\eqalign{ & \eta = \frac{{\left( {627 + 273} \right) - \left( {273 + 27} \right)}}{{627 + 273}} \cr & = \frac{{900 - 300}}{{900}} = \frac{{600}}{{900}} = \frac{2}{3} \cr & {\text{work}} = \left( \eta \right) \times {\text{Heat}} = \frac{2}{3} \times 3 \times {10^6} \times 4.2\,J \cr & = 8.4 \times {10^6}\,J \cr} $$

If a process is expansion then work done is positive so answer will be (A).

$$\eqalign{ & {W_{AB}} = 0,{W_{BC}} = P\Delta V = nR\Delta T = - nR{T_0} \cr & {W_{CA}} = nRT\ell n\frac{{{V_f}}}{{{V_i}}} = nR\left( {2{T_0}} \right)\ell n2 \cr & {Q_{BC}} = n{C_p}\Delta T = \left( {\frac{{nR\gamma }}{{\gamma - 1}}} \right){T_0} \cr & {\text{Efficiency,}}\,\eta = \frac{W}{Q} = \left[ {\frac{{2\ell n2 - 1}}{{\frac{\gamma }{{\left( {\gamma - 1} \right)}}}}} \right] \cr} $$

(A) It is the process by which heat is transmitted from one place to another without heating the intervening medium. Hence, it is not a reversible process.
(B) Nichrome wire is made of alloy and has high resistance. When current is passed through it, heat is produced. So, here electrical energy is converted into heat energy. Hence, it is not a reversible process.
(C) It is the process by which heat is transmitted from one point to another through a substance in the direction of fall of temperature without the actual movement of the particles of the substances themselves. Hence, it cannot be reversible.
(D) Isothermal compression is reversible, for example-Carnot cycle, Heat engine. Thus, choice (D) is correct.

As we know that $${C_p} - {C_V} = R$$
$$\eqalign{ & {C_p} = R + {C_V} \cr & {\text{and}}\,\,\frac{{{C_p}}}{{{C_V}}} = \gamma \,\,\left( {{\text{given}}} \right) \cr & {\text{So,}}\,\,\frac{{R + {C_V}}}{{{C_V}}} = \gamma \cr & \Rightarrow \gamma {C_V} = R + {C_V} \cr & \Rightarrow \gamma {C_V} - {C_V} = R \cr & \Rightarrow {C_V} = \frac{R}{{\gamma - 1}} \cr} $$

$$\eqalign{ & \frac{{{n_1} + {n_2}}}{{\gamma - 1}} = \frac{{{n_1}}}{{{\gamma _1} - 1}} + \frac{{{n_2}}}{{{\gamma _2} - 1}} \cr & {\text{Here, }}{n_1} = {n_2} = 1,{\gamma _1} = \frac{2}{5},{\gamma _2} = \frac{5}{3} \cr} $$

$$\eqalign{ & {T_1}{V^{\gamma - 1}} = {T_2}{\left( {32\,V} \right)^{\gamma - 1}} \cr & \Rightarrow \,\,{T_1} = {\left( {32} \right)^{\gamma - 1}}.{T_2} \cr} $$
For diatomic gas, $$\gamma = \frac{7}{5}$$
$$\eqalign{ & \therefore \,\,\gamma - 1 = \frac{2}{5} \cr & \therefore \,\,{T_1} = {\left( {32} \right)^{\frac{2}{5}}}.{T_2} \cr & \Rightarrow \,\,{T_1} = 4{T_2} \cr} $$
Now, efficiency $$ = 1 - \frac{{{T_2}}}{{{T_1}}}$$
$$\eqalign{ & = 1 - \frac{{{T_2}}}{{4{T_2}}} \cr & = 1 - \frac{1}{4} \cr & = \frac{3}{4} \cr & = 0.75. \cr} $$

The force is $$\left[ {\left( {{P_0} + \frac{{h\rho g}}{2}} \right) \times \left( {2\,R \times h} \right)} \right] - 2\,RT$$

Note : In the first part the force is created due to pressure and in the second part the force is due to surface tension $$T.$$
∴ Force $$ = 2\,{P_0}Rh + R\rho g{h^2} - 2\,RT$$