If 10^{23} molecules are removed from 5.6 L | vedclass

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(D) Step $1$: Calculate the initial number of moles $(n_{initial})$ using the ideal gas equation $PV = nRT$.Given: $P = 2 \ atm$,$V = 5.6 \ L$,$T = 27

Vedclass · Accepted Answer

$1/3$

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(D) Step $1$: Calculate the initial number of moles $(n_{initial})$ using the ideal gas equation $PV = nRT$.Given: $P = 2 \ atm$,$V = 5.6 \ L$,$T = 273 \ K$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.$n_{initial} = \frac{PV}{RT} = \frac{2 	imes 5.6}{0.0821 	imes 273} \approx \frac{11.2}{22.4} = 0.5 \ mol$.Step $2$: Calculate the number of moles removed $(n_{removed})$.$n_{removed} = \frac{	ext{Number of molecules}}{N_A} = \frac{10^{23}}{6.022 	imes 10^{23}} \approx \frac{1}{6} \ mol$.Step $3$: Calculate the remaining moles $(n_{remaining} = n_{initial} - n_{removed})$.$n_{remaining} = 0.5 - \frac{1}{6} = \frac{1}{2} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3} \ mol$.

If $10^{23}$ molecules are removed from $5.6 \ L$ of an ideal gas at $273 \ K$ and $2 \ atm$ pressure,how many moles of the gas will remain?

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