A sample of ideal gas with initial pressure p and volume V is taken through an isothermal expansion proceed during which the change in entropy is found to be ΔS. The universal gas constant is R. Then the work done by the gas is given by

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