Adiabatic process condition

For an adiabatic process, $Q=0$

Using the first law of thermodynamics, $\Delta Q = \Delta U + W$ where $W$ is the work done by the gas.

Since $Q=0$, $0=\Delta U + W \implies W=-\Delta U$

In magnitude, the work done is equal to the change in internal energy. Thus statement (B) is taken as correct in the usual exam convention.

Check statement (A): Internal energy is constant

For a monoatomic ideal gas, $U=\frac{3}{2}nRT$ So internal energy depends only on temperature. In an adiabatic process, temperature generally changes, so internal energy is not constant.

Hence (A) is false.

Check statement (C): Product of temperature and volume is constant

For an adiabatic process of an ideal gas, $PV^\gamma=\text{constant}$ and also $TV^{\gamma-1}=\text{constant}$

For a monoatomic gas, $\gamma=\frac{5}{3}$ so $TV^{2/3}=\text{constant}$ not $TV=\text{constant}$.

Hence (C) is false.

Check statement (D): Product of pressure and volume is constant

$PV=\text{constant}$ is true for an isothermal process, not adiabatic.

In adiabatic process, $PV^\gamma=\text{constant}$

Hence (D) is false.

Check statement (E): Work done to change temperature from $T_1$ to $T_2$ is proportional to $(T_2-T_1)$

For a monoatomic ideal gas, $\Delta U = nC_V(T_2-T_1)$ where $C_V=\frac{3R}{2}$

In an adiabatic process, $W=-\Delta U=-nC_V(T_2-T_1)$

Therefore work done is indeed proportional to $(T_2-T_1)$.

Hence (E) is true.

Conclusion

Correct statements are: $\boxed{(B), (E)}$

Therefore the correct option is $\boxed{\text{B}}$