Thermodynamics and Statistical Physics MCQs Question and Answer in Engineering Physics

A. $$\frac{3}{2}NkT$$

B. $$kT\ln \frac{{\hbar \omega }}{{kT}}$$

C. $$NkT\ln \frac{{\hbar \omega }}{{kT}}$$

D. $$NkT\ln \frac{{\hbar \omega }}{{2kT}}$$

A. $$S = \frac{4}{3}aV{T^3},\,p = \frac{a}{3}{T^4}$$

B. $$S = \frac{1}{3}aV{T^4},\,p = \frac{{4a}}{3}{T^3}$$

C. $$S = \frac{4}{3}aV{T^4},\,p = \frac{a}{3}{T^3}$$

D. $$S = \frac{1}{3}aV{T^3},\,p = \frac{{4a}}{3}{T^4}$$

A. √E

B. E

C. $${{\text{E}}^{\frac{3}{2}}}$$

D. E²

A. Entropy decreases markedly on cooling a superconductor, below the critical, temperature T_C

B. The electronic contribution to the heat capacity in the superconducting state has an exponential form with an argument proportional to T^-1, suggestive of an energy gap

C. A type I superconductor is a perfect diamagnet

D. Critical temperature of superconductors varies with the isotropic mass

A. 96.2%

B. 62.3%

C. 3.5%

D. 37.7%

A. 800 K

B. 700 K

C. 600 K

D. 500 K

A. $$n = \frac{{{{\left( {2m{E_F}} \right)}^{\frac{3}{2}}}}}{{3{\pi ^2}{\hbar ^3}}}$$

B. $$n = \frac{{m{E_F}}}{{\pi {\hbar ^2}}}$$

C. $$n = \frac{{m{E_F}}}{{2\pi {\hbar ^2}}}$$

D. $$n = \frac{{{2^{\frac{3}{2}}}{{\left( {m{E_F}} \right)}^{\frac{1}{2}}}}}{{\pi \hbar }}$$

A. the total entropy of the final system is the same as the sum of the entropy of the individual system always

B. the entropy of mixing is 2Nk_B In 2

C. the entropy of the final system is less than that of sum of the initial entropies of the two gases

D. the entropy of mixing is non-zero when the atoms A and B are of the same type

A. $${T_1} + \frac{a}{R}\left( {\frac{1}{{{V_2}}} + \frac{1}{{{V_1}}}} \right)$$

B. $${T_1} - \frac{2}{3}\frac{a}{R}\left( {\frac{1}{{{V_2}}} - \frac{1}{{{V_1}}}} \right)$$

C. $${T_1} + \frac{2}{3}\frac{a}{R}\left( {\frac{1}{{{V_2}}} - \frac{1}{{{V_1}}}} \right)$$

D. $${T_1} - \frac{1}{3}\frac{a}{R}\left( {\frac{1}{{{V_2}}} - \frac{1}{{{V_1}}}} \right)$$

A. $$\varepsilon \left[ {\frac{{4\exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + 8\exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}{{1 + 2\exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + 4\exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}} \right]$$

B. $$\varepsilon \left[ {\frac{{2\exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + 8\exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}{{1 + 2\exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + 4\exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}} \right]$$

C. $$\varepsilon {\left[ {\frac{{2\exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + 4\exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}{{1 + 2\exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + 4\exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}} \right]^2}$$

D. $$\varepsilon \left[ {\frac{{\exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + 2\exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}{{1 + \exp \left( {\frac{{ - \varepsilon }}{{{k_B}T}}} \right) + \exp \left( {\frac{{ - 2\varepsilon }}{{{k_B}T}}} \right)}}} \right]$$