10 points for best answer and I will love you forever and EVER!
What pressure will a mixture of 1.25 mole of methane, CH4, and 0.575 mole
of propane, C3H8, exert with a volume of 13.0 L at 50 degree's C?
What pressure will a mixture of 1.25 mole of methane, CH4, and 0.575 mole
of propane, C3H8, exert with a volume of 13.0 L at 50 degree's C?
-
Remember that a gas (ANY gas) will exert one atmosphere of pressure at 0°C in a 22.4L container. 22.4L is the number to remember.
You have 1.25 + 0.575 = 1.825 mole, so that would be 1.825 atmospheres at 0° in 22.4L.
But, the volume is 13L, so that's 1.825 · 22.4 ÷ 13 ≈ 3.145 atmospheres at 0°.
But, the temp. is 50°C, not 0°. Convert to °K by adding 273°. That's 323°K rather than 273°K, so that's ~3.146 · 323 ÷ 273 ≈ 3.72 atmospheres (rounded to 3 significant digits to match the input)
You have 1.25 + 0.575 = 1.825 mole, so that would be 1.825 atmospheres at 0° in 22.4L.
But, the volume is 13L, so that's 1.825 · 22.4 ÷ 13 ≈ 3.145 atmospheres at 0°.
But, the temp. is 50°C, not 0°. Convert to °K by adding 273°. That's 323°K rather than 273°K, so that's ~3.146 · 323 ÷ 273 ≈ 3.72 atmospheres (rounded to 3 significant digits to match the input)