Use the first law of thermodynamics for an adiabatic process:

$Q = 0$

So,

$\Delta Q = \Delta U + W = 0$

Hence,

$W = -\Delta U$

where $W$ is the work done on the gas in one sign convention, or equivalently work done by the gas is equal to the decrease in internal energy. The key point is that the magnitude of work depends only on the change in internal energy.

For an ideal gas, internal energy depends only on temperature:

$U = n C_V T$

Therefore,

$\Delta U = n C_V \Delta T$

Substituting into the adiabatic relation:

$W = -n C_V \Delta T$

or, for work done by the gas,

$W = n C_V (T_1 - T_2)$

Thus, for a given gas, the work done in an adiabatic change depends only on the change in temperature.

Check the options:

• A: change in its pressure — No, not only pressure.
• B: change in its temperature — Yes.
• C: change in its specific heat — No, specific heat is a property of the gas, not the changing state variable asked here.
• D: change in its volume — No, not only volume.

Final answer:

$\boxed{\text{B: change in its temperature}}$