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(A) The most probable speed of gas molecules is given by $v_p = \sqrt{\frac{2RT}{M}}$,which implies $v_p \propto \sqrt{T}$.Isotherms represent lines o

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$v_p$ at $3 = v_p$ at $4 > v_p$ at $2 > v_p$ at $1$

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(A) The most probable speed of gas molecules is given by $v_p = \sqrt{\frac{2RT}{M}}$,which implies $v_p \propto \sqrt{T}$.Isotherms represent lines of constant temperature. In a $P-V$ diagram,isotherms further from the origin correspond to higher temperatures.From the figure,the isotherm $EF$ is the furthest from the origin,followed by $CD$,and then $AB$ is the closest.Points $3$ and $4$ lie on the isotherm $EF$,so $T_3 = T_4$. Thus,$v_p$ at $3 = v_p$ at $4$.Point $2$ lies on the isotherm $CD$,so $T_2$ is the temperature of $CD$.Point $1$ lies on the isotherm $AB$,so $T_1$ is the temperature of $AB$.Since the order of temperatures is $T_3 = T_4 > T_2 > T_1$,the order of most probable speeds is $v_p$ at $3 = v_p$ at $4 > v_p$ at $2 > v_p$ at $1$.

$A$ certain amount of an ideal gas is taken from its initial state $1$ to final state $4$ through the path $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ as shown in the figure. $AB, CD, EF$ are all isotherms. If $v_p$ is the most probable speed of the molecules,then:

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