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 = 1.93 × 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 = 2.62 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 = [(0.6p_i x 2.62 x 10^-3 m³) / (p_i x 1.93 x 10^-3m³)](393 K)

T_f = [(0.6 x 2.78 x 10^-3 m³) / (1.93 x 10^-3 m³)](393 K)

T_f = 320.1 K

T_f = (320.1 - 273)^oC

T_f = 47.1^oC

Real gases can approach the behavior of ideal gases under certain conditions. Helium is thought to be the closest to an ideal gas with γ = 1.660. Calculate the molar heat capacities i) CV and ii) Cp of helium.

Real gases can approach the behavior of ideal gases under certain conditions. Helium is thought to be the closest to an ideal gas with γ = 1.660. Calculate the molar heat capacities i) C_V and ii) C_p of helium.