What will be the value of the gas constant R | vedclass

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(D) We know that the gas constant $R = 8.314 \, J \, K^{-1} mol^{-1} = 8.314 \, N \cdot m \cdot K^{-1} mol^{-1}$.Since $1 \, N = 10^5 \, dyne$ and $1

Vedclass · Accepted Answer

$8.314 	imes 10^{14} \, (dyne \, m^{-2})(mm^3) K^{-1} mol^{-1}$

Vedclass · Answer

(D) We know that the gas constant $R = 8.314 \, J \, K^{-1} mol^{-1} = 8.314 \, N \cdot m \cdot K^{-1} mol^{-1}$.Since $1 \, N = 10^5 \, dyne$ and $1 \, m = 10^3 \, mm$,we have $1 \, m^3 = (10^3 \, mm)^3 = 10^9 \, mm^3$.However,the pressure unit given is $dyne \, m^{-2}$. Note that $1 \, N \cdot m = 10^5 \, dyne \cdot m$.To convert $R$ to the required units:$R = 8.314 \, N \cdot m \cdot K^{-1} mol^{-1} = 8.314 	imes (10^5 \, dyne) 	imes (10^9 \, mm^3) \cdot m^{-2} \cdot K^{-1} mol^{-1}$ is incorrect based on the unit provided.Correct conversion: $1 \, J = 10^7 \, erg = 10^7 \, dyne \cdot cm$.Given pressure $P$ in $dyne \cdot m^{-2}$ and volume $V$ in $mm^3$:$R = 8.314 \, J = 8.314 	imes 10^7 \, dyne \cdot cm = 8.314 	imes 10^7 	imes (10^{-2} \, m) 	imes (10^3 \, mm)^3 	imes (10^{-6} \, m^2)^{-1} \dots$Actually,$1 \, J = 10^7 \, dyne \cdot cm = 10^7 \, dyne \cdot (10^{-2} \, m) = 10^5 \, dyne \cdot m$.$1 \, m^3 = 10^9 \, mm^3$,so $1 \, m = 10^3 \, mm$.$R = 8.314 \, N \cdot m = 8.314 	imes (10^5 \, dyne) 	imes (10^9 \, mm^3) 	imes (10^{-3} \, m^{-1}) \dots$ Simplifying: $R = 8.314 	imes 10^5 \, dyne \cdot m \cdot K^{-1} mol^{-1} = 8.314 	imes 10^5 	imes (10^9 \, mm^3) 	imes (10^{-2} \, m^{-2}) = 8.314 	imes 10^{14} \, (dyne \cdot m^{-2})(mm^3) K^{-1} mol^{-1}$.

What will be the value of the gas constant $R$ when pressure is expressed in $dyne \, m^{-2}$ and volume in $mm^3$?

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