14Nov 2018 by Vincent Summers 1 Comment

Pressure Cookers Cook Hotter and Quicker – The Ideal Gas Law

Chemistry, Food
[caption id="attachment_22958" align="alignright" width="480"] Modern 8.5 quart steel pressure cooker[/caption] Your time is limited. Besides, you are concerned about tenderness and nutrition. So you are in the market for a pressure cooker. But you are just a bit curious about how and why they work. And you are not certain which one to buy. Meats and Vegetables Water ordinarily boils at 212° Fahrenheit (100° Celsius). So food cooked in an open saucepan,1 whether by boiling or steaming, cooks at approximately that temperature. Using a Pressure Cooker In an open saucepan, water boils at atmospheric pressure, something over 14 pounds per square inch. However, when water boils in a pressure cooker, steam pressure is considerably higher. Now steam is essentially gaseous water. This brings to mind our high-school days and the…
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29Jan 2018 by Vincent Summers 1 Comment

Converting Gas Temperature to Particle Velocity

Chemistry, Physics
[caption id="attachment_18761" align="alignright" width="440"] Atoms or Molecules of Gas[/caption] Unlike liquids or solids, whose atoms or molecules have greater correlation with each other, gas atoms and molecules move somewhat independently of each other. This is to be expected. A gas occupies a much greater volume than a corresponding liquid. This independent behavior allows us to calculate the root-mean-square velocity of gas particles directly from temperature. Tweaking the Ideal Gas Law We derive this equation from another well-known equation, the Ideal Gas Law equation. Algebraically, that equation is written: PV = nRT where P is pressure, V is volume (not velocity), n is the number of moles of gas, R is the ideal gas constant and T is the temperature. By combining the above equation with derivatives of Boltzmann's equation, we…
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03Mar 2014 by Vincent Summers No Comments

Propane and Oxygen Combustion Question

Chemistry
[caption id="attachment_16612" align="alignright" width="480"] Propane[/caption] Problem: We desire to learn how much oxygen is needed to completely consume a certain quantity of propane gas. Our hydrocarbon and oxygen combustion question follows the basic reaction path, C3H8 + 5 O2 → 4 H2O + 3 CO2 If we have the following conditions: Temperature = 75 Celsius (348 Kelvin) Pressure = 720 / 760 mm = 0.95 atm Moles propane = 40.8 grams / 44.1 grams molecular weight = 0.93 moles How Much Oxygen to Burn the Propane? What volume of oxygen is needed to accomplish the burning of the 0.93 moles of the hydrocarbon? 5 times 0.93 moles of C3H8 burned = 4.65 moles of oxygen The ideal gas law reads: PV = nRT where P= the pressure, V= the volume, and…
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