For 1 mol of an ideal gas,an isochore is obt | vedclass

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(C) For an ideal gas,the equation is $PV = nRT$. For an isochore (constant volume process),the equation can be written as $P = (\frac{nR}{V})T$. The s

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(C) For an ideal gas,the equation is $PV = nRT$. For an isochore (constant volume process),the equation can be written as $P = (\frac{nR}{V})T$. The slope of the $P$ vs $T$ graph is given by $m = \frac{nR}{V}$. Given $n = 1 \ mol$ and slope $m = 0.082 \ atm \ K^{-1}$. So,$\frac{1 	imes 0.082}{V} = 0.082$,which implies $V = 1 \ L$. Now,at $T = 12.2 \ K$,the pressure $P$ is calculated as $P = \frac{nRT}{V} = \frac{1 	imes 0.082 	imes 12.2}{1} = 1.0004 \ atm \approx 1.0 \ atm$.

For $1 \ mol$ of an ideal gas,an isochore is obtained. The slope of the isochore is $0.082 \ atm \ K^{-1}$. What will be its pressure (in $atm$) when the temperature is $12.2 \ K$? $(R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1})$.

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