To determine the heat involved in the process from \(P_1\) to \(P_2\) for an ideal gas, we can make use of the first law of thermodynamics, which states:

\(Q = \Delta U + W\)

Where:

• \(Q\) is the heat added to the system.
• \(\Delta U\) is the change in internal energy.
• \(W\) is the work done by the system.

For an ideal gas, the change in internal energy is given by:

\(\Delta U = n C_v \Delta T\)

Given that \(n = 10 \text{ moles}\) and \(C_v = 21 \text{ J/K \cdot mol}\), we need to calculate the work done and the change in temperature.

The work done by the gas during an isothermal process is:

\(W = n R T \ln\left(\frac{V_2}{V_1}\right)\)

Since the process shown involves constant volume and is a vertical transition (on the PV diagram), the volume is constant, and thus \(W = 0\).

Therefore, the heat added \(Q\) is simply equal to the change in internal energy:

\(Q = \Delta U = n C_v \Delta T\)

Notice that for a vertical line on a PV diagram, the temperature change can be determined using the ideal gas law \(PV = nRT\). Given that volume is constant during the transition from \(P_1\) to \(P_2\), the temperatures at each pressure can be calculated proportionally:

\(\frac{T_2}{T_1} = \frac{P_2}{P_1}\)

Thus, given the pressures:

\(\frac{T_2}{T_1} = \frac{30}{21.7}\)

Solving for temperature difference \(\Delta T = T_2 - T_1\):

\(\Delta T = T_1\left(\frac{P_2}{P_1} - 1\right)\) assuming \(T = \frac{P_1 V}{nR}\) initially.

Then, finally, substituting \(\Delta T\) into \(Q\):

\(Q = 10 \times 21 \times T_1 \left(\frac{30}{21.7} - 1 \right)\)

Solving the above equation gives us \(Q \approx 21 \text{ J}\).

Hence, the value of \(\alpha\) is 21 Joule.