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**Hint**To solve this question, we need to use the first law of thermodynamics. For that, we have to determine the work done by using its expression for the isobaric process. Then substituting the value of the work done, and the given value of the heat, we will get the value of the change in the internal energy.

**Formula used:**The formulae used to solve this question are

${W_p} = P\Delta V$, here ${W_p}$ is the work done in an isobaric process by a pressure of $P$ which changes the volume by $\Delta V$.

$Q = \Delta U + W$, here $Q$ is the heat, $\Delta U$ is the change in the internal energy, and $W$ is the work done.

**Complete step-by-step solution:**

Since the pressure acting on the gas is constant, so the gas undergoes an isobaric process in which the work done is given by

${W_p} = P\Delta V$

$ \Rightarrow {W_p} = P\left( {{V_2} - {V_1}} \right)$

According to the question, $P = 4.5 \times {10^5}Pa$, ${V_1} = 0.5{m^3}$ and ${V_2} = 2.0{m^3}$. Substituting these in above, we get

\[{W_p} = 4.5 \times {10^5}\left( {2.0 - 0.5} \right)\]

$ \Rightarrow {W_p} = 4.5 \times {10^5} \times 1.5$

On solving we get

${W_p} = 6.75 \times {10^5}J$

The heat given to the gas is equal to $800KJ$. So we have $Q = 800KJ$.

We know that $1KJ = {10^3}J$. So we get

$Q = 800 \times {10^3}J$

$ \Rightarrow Q = 8 \times {10^5}J$

From the first law of thermodynamics, we have

$Q = \Delta U + W$

Substituting $Q = 8 \times {10^5}J$ and $W = {W_p} = 6.75 \times {10^5}J$ in the above equation, we get

$8 \times {10^5} = \Delta U + 6.75 \times {10^5}$

$ \Rightarrow \Delta U = 8 \times {10^5} - 6.75 \times {10^5}$

On solving we finally get

$\Delta U = 1.25 \times {10^5}J$

Thus, the change in internal energy of the gas is equal to $1.25 \times {10^5}J$.

**Hence, the correct answer is option D.**

**Note:**The sign convention for the heat must be carefully followed. Since the system is receiving the heat, it is taken as positive. Also, do not forget to write the value of the heat in Joules, which is given in kiloJoules.

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