Question

One mole of an ideal gas is taken through an adiabatic process where the temperature rises from $27^\circ\text{C}$ to $37^\circ\text{C}$. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true?
[R = 8.314 J mol$^{-1}$ K$^{-1}$]

• A. work done by the gas is close to 582 J
• B. work done on the gas is close to 582 J
• C. work done by the gas is close to 332 J
• D. work done on the gas is close to 332 J

Hint:

Remember that for vibrational modes, each mode contributes 2 degrees of freedom (one for kinetic energy and one for potential energy). Also, a negative sign for work in thermodynamics implies work is done ON the system.

Answer 1

The Correct Option is B

For an adiabatic process, the work done is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
First, we need to calculate the degrees of freedom (f) for the polyatomic molecule.
A polyatomic molecule has 3 translational and 3 rotational degrees of freedom.
Each vibrational mode contributes 2 degrees of freedom.
Vibrational degrees of freedom = $4 \times 2 = 8$.
Total degrees of freedom, $f = f_{trans} + f_{rot} + f_{vib} = 3 + 3 + 8 = 14$.
Next, calculate the adiabatic index, $\gamma$.
$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{14} = 1 + \frac{1}{7} = \frac{8}{7}$.
Now, we can calculate the work done using the given values:
n = 1 mole.
R = 8.314 J mol$^{-1}$ K$^{-1}$.
Initial temperature $T_1 = 27^\circ\text{C} = 27 + 273.15 = 300.15 \text{ K}$.
Final temperature $T_2 = 37^\circ\text{C} = 37 + 273.15 = 310.15 \text{ K}$.
$\Delta T = T_2 - T_1 = 10 \text{ K}$, so $T_1 - T_2 = -10 \text{ K}$.
$W = \frac{1 \times 8.314 \times (-10)}{\frac{8}{7} - 1} = \frac{-83.14}{1/7}$.
$W = -83.14 \times 7 = -581.98 \text{ J}$.
The magnitude of the work done is approximately 582 J.
The negative sign indicates that work is done on the gas (compression), which is consistent with the temperature increase in an adiabatic process.
Therefore, the work done on the gas is close to 582 J.