Correct answer is (A)

To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.

First, let's calculate the heat added ($Q$) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:

$Q=m\times L$

where $m=1\mathrm{kg}$ and $L=2257\mathrm{kJ}/\mathrm{kg}$:

$Q=1\mathrm{kg}\times 2257\mathrm{kJ}/\mathrm{kg}=2257\mathrm{kJ}$

Next, let's calculate the work done ($W$) on the system during the process. The work done is given by:

$W=-P\mathrm{\Delta }V$

where $P$ is the atmospheric pressure and $\mathrm{\Delta }V$ is the change in volume. We are given that the atmospheric pressure is $P=1\times {10}^5\mathrm{Pa}$, and the change in volume is

$\mathrm{\Delta }V=1.671{\mathrm{m}}^3-1.00\times {10}^{-3}{\mathrm{m}}^3=1.670{\mathrm{m}}^3$.

Now, we can calculate the work done:

$W=-(1\times {10}^5\mathrm{Pa})(1.670{\mathrm{m}}^3)=-167000\mathrm{J}=-167\mathrm{kJ}$

Finally, we can find the change in internal energy ($\mathrm{\Delta }U$) using the first law of thermodynamics:

$\mathrm{\Delta }U=Q+W$

Substitute the values of $Q$ and $W$:

$\mathrm{\Delta }U=2257\mathrm{kJ}-167\mathrm{kJ}=2090\mathrm{kJ}$

The change in internal energy of the system during the process is +2090 kJ.

Correct answer is (B)

Statement I: If heat is added to a system, its temperature must increase. This statement is not necessarily true. For example, in a phase transition (like melting or boiling), heat can be added to a system without increasing its temperature. The added heat energy is used to break intermolecular bonds and change the phase of the substance, not to increase the kinetic energy of the particles (which would raise the temperature).

Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase. This statement is generally true, as positive work being done by a system often involves expansion against an external pressure, thus increasing its volume.

Correct answer is (D)

The rate of increase of internal energy of a system can be found using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

In this case, the heat being supplied to the system is 1000 W and the work being done by the system is 200 W.

Therefore, the rate at which the internal energy of the system increases is:

1000 W (heat supplied) - 200 W (work done) = 800 W

So, the correct answer is 800 W.

Correct answer is (B)

$\because \mathrm{\Delta }Q=0$

$\begin{array}{rl}\mathrm{\Delta }U& =-W\\ \\ & =-[-\frac{1}{2}\times (50+10)\times {10}^3\times 150\times {10}^{-6}]\\ \\ & =4.5\mathrm{J}\end{array}$

Correct answer is (B)

Heat supplied to the system $=dQ$

Work done on the system $=-dW$

By the first law of thermodynamics, $dQ=dU+dW$

We done the work, therefore,

$dQ=dU+(-dW)=dU-dW$

So, both A and R are correct and R is the correct explanation of A.