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If 77.0 mL of nitrogen gas is collected over water at 50 °C and 763 mm Hg, what is the volume of dry nitrogen gas at STP? The vapor pressure of water at 50 °C is 92.5mm Hg.

Sagot :

The volume of the dry nitrogen gas at STP is 57.4 mL

From the question given above, the following data were obtained:

Initial volume (V₁) = 77 mL

Initial temperature (T₁) = 50 °C = 50 + 273 = 323 K

Initial pressure (P₁) = 763 – 92.5 = 670.5 mmHg

Final temperature (T₂) = STP = 273 K

Final pressure (P₂) = 760 mmHg

Final volume (V₂) =?

The volume of the dry nitrogen can be obtained by using the combine gas equation as follow:

[tex] \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\ \\ \frac{670.5 \times 77}{323} = \frac{760 \times V_2}{273} \\ \\ cross \: multiply \\ \\ 323 \times 760 \times V_2 = 670.5 \times 77 \times 273 \\ \\ divide \: both \: side \: by \: 323 \times 760 \\ \\ V_2 = \: \frac{670.5 \times 77 \times 273 }{323 \times 760} \\ \\ V_2 = 57.4 \: ml[/tex]

Therefore, the volume of the dry nitrogen gas is 57.4 mL

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