10 mole of an ideal gas is undergoing the process shown in the...

From the given $P-V$ diagram, the process from $P_1$ to $P_2$ is an isochoric process because the volume remains constant at $V = 1\text{ m}^3$.
For an isochoric process, the work done $W = 0$.
According to the first law of thermodynamics, the heat involved is $Q = \Delta U + W = \Delta U$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
Using the ideal gas equation $PV = nRT$, we have $n\Delta T = \frac{V\Delta P}{R}$.
Substituting this into the expression for $Q$:
$Q = C_v \left( \frac{V(P_2 - P_1)}{R} \right)$
Given values: $n = 10$ moles, $V = 1\text{ m}^3$, $P_1 = 21.7\text{ Pa}$, $P_2 = 30\text{ Pa}$, $C_v = 21\text{ J/mol}\cdot\text{K}$, and $R = 8.3\text{ J/mol}\cdot\text{K}$.
$Q = 21 \times \frac{1 \times (30 - 21.7)}{8.3}$
$Q = 21 \times \frac{8.3}{8.3} = 21\text{ J}$.
Thus, the value of $\alpha$ is 21.