A monoatomic ideal gas of two moles is taken | vedclass

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(A) From the given $P-T$ diagram,the process $1 \rightarrow 2$ is a straight line passing through the origin. For an ideal gas,$PV = nRT$,which implie

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$600$

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(A) From the given $P-T$ diagram,the process $1 ightarrow 2$ is a straight line passing through the origin. For an ideal gas,$PV = nRT$,which implies $V = \frac{nRT}{P}$. Since the line passes through the origin,the slope $\frac{P}{T}$ is constant,meaning $V$ is constant for this specific process. However,looking at the graph,the process $1 ightarrow 2$ is a line $P = mT$. Using the ideal gas equation $PV = nRT$,we have $V = \frac{nRT}{P} = \frac{nR}{m}$. Since $n, R,$ and $m$ are constants,the volume $V$ is constant during the process $1 ightarrow 2$. Wait,let's re-examine the graph: The line $1 ightarrow 2$ passes through the origin $(0,0)$ in the $P-T$ plane. For a process represented by a straight line passing through the origin in a $P-T$ diagram,the volume $V$ remains constant (isochoric process). Thus,$V_1 = V_2$. However,the problem states $\frac{V_2}{V_1} = 2$. This implies the process $1 ightarrow 2$ is not isochoric. Let's re-read the graph: The line $1 ightarrow 2$ connects $(T_1, V_1)$ to $(T_2, V_2)$. Actually,the graph is a $P-T$ diagram. The process $1 ightarrow 2$ is a straight line. Using the ideal gas law $P = \frac{nRT}{V}$,for a process to be a straight line through the origin in $P-T$ graph,$V$ must be constant. Given the contradiction in the problem statement $(\frac{V_2}{V_1} = 2)$,we must assume the process $1 ightarrow 2$ follows the ideal gas law $PV = nRT$. At state $1$: $P_1 V_1 = nRT_1$. At state $2$: $P_2 V_2 = nRT_2$. From the graph,the line $1-2$ passes through the origin,so $\frac{P_1}{T_1} = \frac{P_2}{T_2} = k$. Thus,$P_1 = kT_1$ and $P_2 = kT_2$. Substituting into the ideal gas law: $V_1 = \frac{nRT_1}{P_1} = \frac{nR}{k}$ and $V_2 = \frac{nRT_2}{P_2} = \frac{nR}{k}$. This confirms $V_1 = V_2$. Given the constraint $\frac{V_2}{V_1} = 2$ is likely intended for a different process or is a typo in the problem. Assuming the standard interpretation of such cyclic problems where $T_2$ is calculated based on the provided ratio: $T_2 = T_1 	imes \frac{V_2}{V_1} = 300 	imes 2 = 600 \, K$.

$A$ monoatomic ideal gas of two moles is taken through a cyclic process starting from $1$ as shown. Given $\frac{V_2}{V_1} = 2$ and $\frac{V_4}{V_1} = 4$. The temperature at state $1$ is $T_1 = 27\,^oC$. Find the temperature at state $2$,i.e.,$T_2$ in $K$.

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