10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the proc

Question

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 $\alpha$. Joule ( $P_1=21.7 \mathrm{~Pa}$ and $\left.P_2=30 \mathrm{~Pa}, \mathrm{C}_v=21 \mathrm{~J} / \mathrm{K} . \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\right)$. The value of $\alpha$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

By examining the thermodynamic process illustrated in the specific $P-V$ or $P-T$ diagram, we ascertain the nature of the expansion or compression. Using the First Law of Thermodynamics ($Q = \Delta U + W$), we calculate the change in internal energy $\Delta U = n C_v \Delta T$ and the work done $W$. Applying the provided constants ($C_v = 21 	ext{ J/K}\cdot	ext{mol}$ and $R = 8.3 	ext{ J/mol}\cdot	ext{K}$), the computation of the total heat transferred $Q$ yields a value of exactly 21 Joules.