Ten moles of an ideal monoatomic gas, initially in state a at...

For state $a$: $n = 10$, $T_a = 27^{\circ}\text{C} = 300\text{ K}$, $P_a = 1\text{ atm}$, $V_a = V_0$.
For a monoatomic gas, $C_v = \dfrac{3}{2}R$ and $\gamma = \dfrac{5}{3}$.

Process $a \rightarrow b$ is a sudden compression, which is an adiabatic process.
$V_b = \dfrac{V_0}{3}$
Using $T_a V_a^{\gamma-1} = T_b V_b^{\gamma-1}$:
$T_b = T_a \left(\dfrac{V_a}{V_b}\right)^{\gamma-1} = 300 \times (3)^{5/3 - 1} = 300 \times 3^{2/3} = 300 \times (9)^{1/3}$
Given $9^{1/3} = 2.08$, we get $T_b = 300 \times 2.08 = 624\text{ K}$.
Using $P_a V_a^\gamma = P_b V_b^\gamma$:
$P_b = P_a \left(\dfrac{V_a}{V_b}\right)^\gamma = 1 \times (3)^{5/3} = 3 \times 3^{2/3} = 3 \times 2.08 = 6.24\text{ atm}$.
Thus, the pressure at state $b$ is $6.24$ times the atmospheric pressure. Option (D) is incorrect.

The change in internal energy for process $a \rightarrow b$ is:
$\Delta U_{ab} = n C_v (T_b - T_a) = 10 \times \dfrac{3}{2}R \times (624 - 300) = 15R \times 324 = 4860R$.
Option (B) is correct.

Process $b \rightarrow c$ is isochoric (stationary piston), so $V_c = \dfrac{V_0}{3}$.
The gas cools to the water bath temperature, $T_c = 11^{\circ}\text{C} = 284\text{ K}$.

Process $c \rightarrow f$ is isothermal (slow expansion in water bath), so $T_f = T_c = 284\text{ K}$ and $V_f = V_0$.
The net change in internal energy for the entire process $a \rightarrow f$ depends only on the initial and final temperatures:
$\Delta U_{net} = n C_v (T_f - T_a) = 10 \times \dfrac{3}{2}R \times (284 - 300) = 15R \times (-16) = -240R$.
Option (C) is correct.

For the P-V diagram:
Process $a \rightarrow b$ is an adiabatic compression (curve upwards and to the left).
Process $b \rightarrow c$ is an isochoric cooling (vertical line downwards).
Process $c \rightarrow f$ is an isothermal expansion (curve downwards and to the right).
Comparing pressures at volume $V_0$:
$P_a = \dfrac{nR(300)}{V_0}$ and $P_f = \dfrac{nR(284)}{V_0}$
Since $P_a > P_f$, point $a$ lies above point $f$. The given schematic diagram correctly represents all these features. Option (A) is correct.

Answer: The schematic P-V diagram of the processes described above is: ; The change in internal energy in going from state $\boldsymbol{a}$ to $\boldsymbol{b}$ is $4860R$.; The net change in the internal energy in the whole process is $-240R$.