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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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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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