Question 664: Thermodynamics & KTG

Question

A polyatomic gas at pressure P , having volume ' $V$ ' expands isothermally to a volume ' 3 V ' and then adiabatically to a volume ' 24 V '. The final pressure of gas is (for moderate temperature changes)

Accepted Answer

Step 1: Isothermal Expansion
For isothermal process:
$ P V = P_1 V_1 $
Initial state: $P, V$.
After isothermal expansion to $3V$:
$ P_i V_i = P_1 V_1 \implies P \cdot V = P_2 \cdot (3V) $
$ P_2 = \frac{P}{3} $
So, after first process:
$ (P_2, V_2) = \left(	frac{P}{3}, 3Vight). $
Step 2: Adiabatic Expansion
Now expand adiabatically from $V_2 = 3V$ to $V_3 = 24V$.
Adiabatic relation:
$ P V^{\gamma} = 	ext{constant}. $
So:
$ P_2 \,(V_2^{\gamma}) = P_3 (V_3^{\gamma}). $
$ P_3 = P_2 \left(\frac{V_2}{V_3}ight)^\gamma. $
Substitute:
$ P_3 = \frac{P}{3} \left(\frac{3V}{24V}ight)^\gamma = \frac{P}{3} \left(\frac{1}{8}ight)^\gamma. $
Step 3: Value of ÃÂ³ for a Polyatomic Gas
For a polyatomic gas, "moderate temperatures" Ã¢ÂÂ assume
non-linear polyatomic molecule
, having $f=6$ degrees of freedom (3 translational + 3 rotational, neglecting vibrations).
$ C_v = \frac{f}{2}R = 3R,\quad C_p = C_v + R = 4R $
$ \gamma = \frac{C_p}{C_v} = 	frac{4R}{3R} = 	frac{4}{3}. $
Step 4: Compute Final Pressure
$ P_3 = \frac{P}{3} \left(\frac{1}{8}ight)^{4/3}. $
Now simplify.
$(1/8)^{4/3} = \left(8^{-1}ight)^{4/3} = 8^{-4/3}.$
Since $8 = 2^3$,
$ 8^{4/3} = (2^3)^{4/3} = 2^4 = 16. $
So:
$ (1/8)^{4/3} = \frac{1}{16}. $
Therefore:
$ P_3 = \frac{P}{3} \cdot \frac{1}{16} = \frac{P}{48}. $
Ã¢ÂÂ Final Answer:
$ \boxed{	frac{P}{48}} $
Correct option: D.

Answer

16 P

Answer

24 P

Answer

P / 36

Answer

P / 48

Thermodynamics & KTG

A polyatomic gas at pressure P , having volume ' V ' expands isothermally to a volume ' 3 V ' and then adiabatically to a volume ' 24 V '. The final pressure of gas is (for moderate temperature changes)

Choose the correct answer: