10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ is $\alpha \text{ Joule}$ ($P_1 = 21.7 \text{ Pa}$ and $P_2 = 30 \text{ Pa}, C_v = 21 \text{ J/K.mol}, R = 8.3 \text{ J/mol.K}$). The value of $\alpha$ is _______.
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:
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.
Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^\circ\text{C}$ and $40^\circ\text{C}$ respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points $B$ and $E$ are (close to):

Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^\circ\text{C}$ and $40^\circ\text{C}$ respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points $B$ and $E$ are (close to):
