The pressure versus temperature diagram of a given system at certain low temperature range is found to be parallel to the temperature axis in the liquid to solid transition region. The change in the specific volume remains constant in this region. The conclusion one can get from the above is

A system has two energy levels with energies E and 2E. The lower level is four-fold degenerate while the upper level is doubly degenerate. If there are N non-interacting classical particles in the system, which is in thermodynamic equilibrium at temperature T, the fraction of particles in the upper level is

A. $$\frac{1}{{1 + {e^{ - \varepsilon /{k_B}T}}}}$$

B. $$\frac{1}{{1 + 2{e^{\varepsilon /{k_B}T}}}}$$

C. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} + 4{e^{2\varepsilon /{k_B}T}}}}$$

D. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} - 4{e^{2\varepsilon /{k_B}T}}}}$$

A system of non-interacting Fermi particles with Fermi energy E_F has the density of states proportional to √E, where E is the energy of a particle. The average energy per particle at temperature T = 0 is

A. $$\frac{1}{6}{E_F}$$

B. $$\frac{1}{5}{E_F}$$

C. $$\frac{2}{5}{E_F}$$

D. $$\frac{3}{5}{E_F}$$

A piston containing an ideal gas is originally in the state X (see figure). The gas is taken through a thermal cycle X → Y → X as shown

The work done by the gas is positive, if the direction of the thermal cycle is

A. clockwise

B. counter-clockwise

C. neither clockwise nor counter-clockwise

D. clockwise from X → Y and counter-clockwise from Y → X

A photon gas is at thermal equilibrium at temperature T. The mean number of photons in an energy state $$\varepsilon = h\omega $$  is

A. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) + 1$$

B. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1$$

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

D. $${\left[ {\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1} \right]^{ - 1}}$$