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(B) According to the ideal gas equation,$p V = n R T$.Rearranging for the volume as a function of temperature,we get $V = (\frac{n R}{p}) T$.The slope

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$0.5$

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(B) According to the ideal gas equation,$p V = n R T$.Rearranging for the volume as a function of temperature,we get $V = (\frac{n R}{p}) T$.The slope of the plot of $V$ versus $T$ is given by $	ext{slope} = \frac{n R}{p}$.Given: $	ext{slope} = 0.328 \, L \, K^{-1}$,$n = 2 \, mol$,and $R = 0.0821 \, L \, atm \, mol^{-1} \, K^{-1}$.Substituting the values: $0.328 = \frac{2 	imes 0.0821}{p}$.Solving for $p$: $p = \frac{2 	imes 0.0821}{0.328} = \frac{0.1642}{0.328} = 0.5 \, atm$.

At a constant pressure $p$,the plot of volume $(V)$ as a function of temperature $(T)$ for $2 \, mol$ of an ideal gas gives a straight line with a slope $0.328 \, L \, K^{-1}$. The value of $p$ (in $atm$) is closest to
[Gas constant,$R=0.0821 \, L \, atm \, mol^{-1} \, K^{-1}$ ]

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At a constant pressure $p$,the plot of volume $(V)$ as a function of temperature $(T)$ for $2 \, mol$ of an ideal gas gives a straight line with a slope $0.328 \, L \, K^{-1}$. The value of $p$ (in $atm$) is closest to[Gas constant,$R=0.0821 \, L \, atm \, mol^{-1} \, K^{-1}$ ]