The ideal gas law says;

PV = nRT

The rearrangement of ideal gas law gives;

(PV) / T = nR

i.e., (PV) / T = constant

Thus, the rearrangement of ideal gas law when mol number is constant leads to;

p_iV_i / T_i= p_fV_f/ T_f

T_f = [(p_fV_f) / (p_iV_i)]T_i ………… (1)

where p_i and p_f are the initial and final pressure, V_i and V_f are the initial and final volume, and T_i and T_fare the initial and final temperatures.

First finding the expression for final pressure;

It is given that pressure drops by 40%, So

p_f= p_i- 40% of p_i

p_f = p_i - (40/100)p_i

p_f = (1 - 0.4)p_i

p_f= 0.6p_i

Now finding the values of the initial and final volume;

The initial volume is given by using the ideal gas law as follows;

V_i = nRT_i / p_i

Substituting values of n, R, T_i and p_i in this equation

V_i= (0.15 mol x 8.31 J/mol.K x 393 K) / (2.5 x 1.013 x 10⁵ Pa)

V_i = 1.934 × 10^-3 m³

Since it is an adiabatic process, the final volume can be found as;

p_iV_i^𝛶 = p_fV_f^𝛶

V_f^𝛶 = (p_i / p_f)V_i^𝛶

V_f = (p_i / p_f)^1/𝛶V_i

Substituting values of p_i, p_f, and V_i in this equation

V_f= (p_i / 0.6p_i)^3/5(1.93 x 10^-3 m³)

V_f = 2.622 x 10^-3 m³

Now substituting values of p_f , V_i , V_f, and T_i in eq(1) to find the final temperature;

T_f = [(p_fV_f) / (p_iV_i)]T_i

T_f = [(0.6p_i x 2.622 x 10^-3 m³) / (p_i x 1.934 x 10^-3m³)](393 K)

T_f = [(0.6 x 2.78 x 10^-3 m³) / (1.934 x 10^-3 m³)](393 K)

T_f = 319.68 K

T_f = (319.68 - 273)^oC

T_f = 46.7^oC

A neon sample undergoes adiabatic expansion from V0 to 2V0, where V0 is the initial volume. Will the mean free path increase/decrease, and by what factor? If not, give a reason.

A neon sample undergoes adiabatic expansion from V₀ to 2V₀, where V₀ is the initial volume. Will the mean free path increase/decrease, and by what factor? If not, give a reason.