2.2 , g of nitrous oxide (N_{2}O) gas is cool | vedclass

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(D) $1$. Calculate the number of moles of $N_{2}O$: $n = \frac{2.2 \, g}{44 \, g \, mol^{-1}} = 0.05 \, mol$.$2$. Calculate the change in enthalpy $(\

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$195$

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(D) $1$. Calculate the number of moles of $N_{2}O$: $n = \frac{2.2 \, g}{44 \, g \, mol^{-1}} = 0.05 \, mol$.$2$. Calculate the change in enthalpy $(\Delta H)$ at constant pressure: $\Delta H = n C_{p} \Delta T = 0.05 \, mol 	imes 100 \, J \, K^{-1} \, mol^{-1} 	imes (270 - 310) \, K = 5 	imes (-40) = -200 \, J$.$3$. Calculate the work done $(w)$ during the process: $w = -P_{ext} \Delta V = -1 \, atm 	imes (167.75 - 217.1) \, mL = -1 	imes (-49.35) \, mL \cdot atm = 49.35 \, mL \cdot atm$.$4$. Convert $w$ to Joules: $1 \, L \cdot atm = 101.3 \, J$,so $w = 49.35 	imes 10^{-3} \, L 	imes 101.3 \, J \, L^{-1} \approx 5 \, J$.$5$. Use the relation $\Delta H = \Delta U + P \Delta V$,which implies $\Delta U = \Delta H - P \Delta V = \Delta H + w$.$6$. $\Delta U = -200 \, J + 5 \, J = -195 \, J$.$7$. Since $\Delta U = -x \, J$,we have $-195 = -x$,so $x = 195$.

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