3.4 moles of an ideal gas occupies a volume o | vedclass

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(A) Given: $n = 3.4 \ mol$,$T = 300 \ K$,$V = 68 \ mL = 68 	imes 10^{-3} \ L = 0.068 \ dm^3$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Using the ideal gas

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$1.247 	imes 10^2 \ kPa$

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(A) Given: $n = 3.4 \ mol$,$T = 300 \ K$,$V = 68 \ mL = 68 	imes 10^{-3} \ L = 0.068 \ dm^3$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Using the ideal gas equation: $PV = nRT$.$P = \frac{nRT}{V} = \frac{3.4 \ mol 	imes 8.314 \ J \ K^{-1} \ mol^{-1} 	imes 300 \ K}{0.068 \ dm^3}$.Since $1 \ J = 1 \ Pa \cdot m^3$ and $1 \ dm^3 = 10^{-3} \ m^3$,the pressure $P = \frac{8480.28}{0.068 	imes 10^{-3}} \ Pa = 124710000 \ Pa = 1.2471 	imes 10^5 \ kPa$ is incorrect in the original prompt's unit conversion.Recalculating: $P = \frac{3.4 	imes 8.314 	imes 300}{0.068 	imes 10^{-3}} \ Pa = 1.2471 	imes 10^8 \ Pa = 1.2471 	imes 10^5 \ kPa$.

$3.4$ moles of an ideal gas occupies a volume of $68 \ mL$ at $300 \ K$. What would be the pressure of the gas? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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