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(A) From the ideal gas equation,$PV = nRT$. For $n = 1$ mole,$T = \frac{PV}{R}$.State $1$: $P_1 = P_0, V_1 = V_0$. Thus,$T_1 = \frac{P_0 V_0}{R} = T_0

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$1 : 2 : 2$

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(A) From the ideal gas equation,$PV = nRT$. For $n = 1$ mole,$T = \frac{PV}{R}$.State $1$: $P_1 = P_0, V_1 = V_0$. Thus,$T_1 = \frac{P_0 V_0}{R} = T_0$.State $2$: $P_2 = 4P_0, V_2 = V_0$. Thus,$T_2 = \frac{4P_0 V_0}{R} = 4T_0$.State $3$: $P_3 = P_0, V_3 = 4V_0$. Thus,$T_3 = \frac{P_0 (4V_0)}{R} = 4T_0$.The average molecular speed $v_{avg}$ is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$,which implies $v_{avg} \propto \sqrt{T}$.Therefore,the ratio of average molecular speeds in states $1, 2$ and $3$ is:$v_1 : v_2 : v_3 = \sqrt{T_1} : \sqrt{T_2} : \sqrt{T_3}$$v_1 : v_2 : v_3 = \sqrt{T_0} : \sqrt{4T_0} : \sqrt{4T_0}$$v_1 : v_2 : v_3 = 1 : 2 : 2$.

One mole of an ideal diatomic gas is taken through the cycle as shown in the figure.
$1 \rightarrow 2$: isochoric process
$2 \rightarrow 3$: straight line on $P-V$ diagram
$3 \rightarrow 1$: isobaric process
The average molecular speed of the gas in the states $1, 2$ and $3$ are in the ratio

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One mole of an ideal diatomic gas is taken through the cycle as shown in the figure.$1 \rightarrow 2$: isochoric process$2 \rightarrow 3$: straight line on $P-V$ diagram$3 \rightarrow 1$: isobaric processThe average molecular speed of the gas in the states $1, 2$ and $3$ are in the ratio