1. Use the first law for an adiabatic process

For an adiabatic process, $Q=0$ So from the first law, $\Delta U = Q - W = -W$ where $W$ is the work done by the gas.

Thus,

• if $\Delta U>0$, then $W<0$
• meaning work is done on the gas.

Here temperature increases from $27^\circ C = 300\,K$ to $37^\circ C = 310\,K$ so $\Delta T = 10\,K$ Hence internal energy increases.

2. Find degrees of freedom of the gas

The gas is polyatomic and has 4 vibrational modes.

For a non-linear polyatomic molecule:

• translational degrees of freedom $=3$
• rotational degrees of freedom $=3$
• each vibrational mode contributes $2$ quadratic degrees of freedom

So total degrees of freedom: $f = 3+3+2\times 4 = 14$

Therefore, $C_V = \frac{f}{2}R = \frac{14}{2}R = 7R$

3. Calculate change in internal energy

For 1 mole ideal gas, $\Delta U = n C_V \Delta T$ With $n=1$, $\Delta U = 1\cdot 7R \cdot 10$ $\Delta U = 70R$ Using $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$, $\Delta U = 70\times 8.314 = 581.98\,\text{J}$ So, $\Delta U \approx 582\,\text{J}$

4. Relate this to work

Since adiabatic, $W_{\text{by gas}} = -\Delta U \approx -582\,\text{J}$ This means the gas does negative work, i.e. 582 J of work is done on the gas.

5. Check options

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

So the correct option is $\boxed{B}$