One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27°C to 37°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$$^{-1}$$]

First, we change the given temperatures into the absolute (kelvin) scale because all thermodynamic formulas use kelvin.

We have $$T_{1}=27^{\circ}\text{C}=27+273=300\ \text{K}$$ and $$T_{2}=37^{\circ}\text{C}=37+273=310\ \text{K}.$$

Now we must find the molar heat capacity at constant volume, $$C_{V},$$ of the gas. According to the classical equipartition theorem every quadratic degree of freedom contributes $$\tfrac12kT$$ (per molecule) to the internal energy. For one mole the contribution becomes $$\tfrac12RT.$$ Hence we first count the quadratic degrees of freedom (d.o.f.).

For a non-linear polyatomic molecule we have

• 3 translational d.o.f.
• 3 rotational d.o.f.
• 4 vibrational modes, and each vibrational mode supplies two quadratic d.o.f. (one kinetic + one potential), so $$4\times2=8$$ vibrational d.o.f.

Therefore the total number of quadratic degrees of freedom is

$$f = 3 + 3 + 8 = 14.$$

Stating the equipartition formula for one mole, the internal energy is

$$U = \frac{f}{2}RT.$$

Consequently the molar heat capacity at constant volume is

$$C_{V} = \frac{\partial U}{\partial T} = \frac{f}{2}R = \frac{14}{2}R = 7R.$$

We now calculate the change in internal energy when the temperature changes by $$\Delta T = T_{2}-T_{1} = 310\ \text{K} - 300\ \text{K} = 10\ \text{K}.$$

Hence (for one mole)

$$\Delta U = C_{V}\,\Delta T = 7R\,(10\ \text{K}).$$

Substituting $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1},$$ we get

$$\Delta U = 7 \times 8.314 \times 10 = 581.98\ \text{J} \approx 582\ \text{J}.$$

The process is specified to be adiabatic, so we write the first law in the form

$$\Delta U = Q - W,$$

and because $$Q = 0$$ for an adiabatic change, this simplifies to

$$\Delta U = -W.$$

Rearranging, we have

$$W = -\Delta U.$$

Since $$\Delta U$$ is positive (the gas has gained internal energy), $$W$$ is negative, meaning that the work is done on the gas. Its magnitude equals $$|\Delta U| \approx 582\ \text{J}.$$

Therefore, the work done on the gas is close to 582 J.

Hence, the correct answer is Option 2.