Question #5609e

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Question #5609e

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Answer 1 · 2015-01-02T09:33:12

I'm assuming that 616 mg is actually 616 mmHg, the gas' pressure.

You can approach this problem by using the combined gas law, since the number of moles does not change.

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$

You know that

$V_{1} = 322$ $V_1 = 322$ $m L$ $"mL$ ,
$T_{1} = 29^{\circ} C = 273.15 + 29 = 302.15$ $T_1 = 29^@C = 273.15 + 29 =302.15$ $K$ $"K"$
$P_{1} = 616$ $P_1 = 616$ $mmHg = 616$ $"mmHg" = 616$ $mmHg \cdot \frac{1 atm}{760 mmHg} = 0.810$ $"mmHg" * ("1 atm")/(760"mmHg") = 0.810$ $atm$ $"atm"$

At STP, the temperature is equal to $T_{2} = 273.15$ $T_2 = 273.15$ $K$ $"K"$ and pressure is equal to $P_{2} = 1.00$ $P_2 = 1.00$ $atm$ $"atm"$ . Plug all in and get

$V_{2} = \frac{P_{1}}{P_{2}} \cdot \frac{T_{2}}{T_{1}} \cdot V_{1} = \frac{0.810 a t m}{1.00 a t m} \cdot \frac{273.15 K}{302.15 K} \cdot 322 m L$ $V_2 = P_1/(P_2) * T_2/T_1 * V_1 = (0.810atm)/(1.00atm) * (273.15K)/(302.15K) * 322mL$

$V_{2} = 236$ $V_2 = 236$ $mL$ $"mL"$

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Question #5609e

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