Question:hard
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume of this process is $TV^x = \text{constant}$. Then $x$ is
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume of this process is $TV^x = \text{constant}$. Then $x$ is
Show Hint
For a monoatomic gas ($f=3$), $x = 2/3$. For a diatomic gas ($f=5$), $x = 2/5$. The value of $x$ is simply $2/f$. Memorizing this ratio saves time!
Updated On: Jul 1, 2026
- $\frac{5}{3}$
- $\frac{2}{5}$
- $\frac{2}{3}$
- $\frac{3}{5}$
Show Solution
The Correct Option is B
Solution and Explanation
1. The Adiabatic Relation: The standard equation is $TV^{\gamma-1} = \text{constant}$.
Comparing this to the given $TV^x = \text{constant}$, we find:
$$x = \gamma - 1$$
2. Determine $\gamma$ for Diatomic Gas: For a rigid diatomic gas (at room temperature, vibrational modes are not active): The degrees of freedom $f = 5$ (3 translational + 2 rotational). $$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}\lt strong\gt 3. Calculate $x$:\lt /strong\gt x = \gamma - 1$$ $$x = \frac{7}{5} - 1 = \frac{2}{5}$$ Therefore, the value of $x$ is $2/5$.
2. Determine $\gamma$ for Diatomic Gas: For a rigid diatomic gas (at room temperature, vibrational modes are not active): The degrees of freedom $f = 5$ (3 translational + 2 rotational). $$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}\lt strong\gt 3. Calculate $x$:\lt /strong\gt x = \gamma - 1$$ $$x = \frac{7}{5} - 1 = \frac{2}{5}$$ Therefore, the value of $x$ is $2/5$.
Was this answer helpful?
0
Top Questions on Thermodynamics
- The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have:
- JEE Main - 2025
- Physics
- Thermodynamics
- Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.
Reason (R): In an isothermal process, \( PV = \text{constant} \), while in an adiabatic process \( PV^\gamma = \text{constant} \). Here, \( \gamma \) is the ratio of specific heats, \( P \) is the pressure and \( V \) is the volume of the ideal gas.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Physics
- Thermodynamics
- A poly-atomic molecule (C\(_3\)R, \(C_v = 4R\), where \(R\) is gas constant) goes from phase space point A (\(P_A = 10^4 \, \text{Pa}, V_A = 4 \times 10^{-3} \, \text{m}^3\)) to point B (\(P_B = 5 \times 10^4 \, \text{Pa}, V_B = 6 \times 10^{-7} \, \text{m}^3\)) to point C (\(P_C = 10^4 \, \text{Pa}, V_C = 8 \times 10^{-3} \, \text{m}^3\)). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:
- JEE Main - 2025
- Physics
- Thermodynamics
- If \[ \text{C(diamond)} \rightarrow \text{C(graphite)} + X \, \text{kj mol}^{-1} \] \[ \text{C(diamond)} + \text{O}_2(g) \rightarrow \text{CO}(g) + Y \, \text{kj mol}^{-1} \] \[ \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}(g) + Z \, \text{kj mol}^{-1} \] At constant temperature. Then:
- JEE Main - 2025
- Chemistry
- Thermodynamics
- Want to practice more? Try solving extra ecology questions todayView All Questions