The Rankine temperature scale used in engineering is to the...

The Rankine temperature scale used in engineering is to the Fahrenheit scale as the Kelvin scale is to the Celsius scale. That is, 1 Rankine degree is the same size as 1 Fahrenheit degree, and $0^{\circ} \mathrm{R}=$ absolute zero. (a) What temperature corresponds to the freezing point of water on the Rankine scale? (b) What is the value of the gas constant $R$ on the Rankine scale in $(\mathrm{L} \cdot \text { atm }) /\left(^{\circ} \mathrm{R} \cdot \text { mol)? }\right.$ (c) Use the van der Waals equation to determine the pressure inside a 400.0 $\mathrm{mL}$ vessel that contains 2.50 $\mathrm{mol}$ of $\mathrm{CH}_{4}$ at a temperature of $525^{\circ} \mathrm{R}$ . For $\mathrm{CH}_{4}, a=2.253$ $\left(\mathrm{L}^{2} \cdot \mathrm{atm}\right) / \mathrm{mol}^{2} ; b=0.04278 \mathrm{L} / \mathrm{mol} .$

The Rankine temperature scale used in engineering is to the Fahrenheit scale as the Kelvin scale is to the Celsius scale. That is, 1 Rankine degree is the same size as 1 Fahrenheit degree, and $0^{\circ} \mathrm{R}=$ absolute zero.
(a) What temperature corresponds to the freezing point of water on the Rankine scale?
(b) What is the value of the gas constant $R$ on the Rankine scale in $(\mathrm{L} \cdot \text { atm }) /\left(^{\circ} \mathrm{R} \cdot \text { mol)? }\right.$
(c) Use the van der Waals equation to determine the pressure inside a 400.0 $\mathrm{mL}$ vessel that contains 2.50 $\mathrm{mol}$ of $\mathrm{CH}_{4}$ at a temperature of $525^{\circ} \mathrm{R}$ . For $\mathrm{CH}_{4}, a=2.253$ $\left(\mathrm{L}^{2} \cdot \mathrm{atm}\right) / \mathrm{mol}^{2} ; b=0.04278 \mathrm{L} / \mathrm{mol} .$

Key Concepts

Temperature Scale Conversion

Temperature scale conversion involves converting measurements from one temperature unit to another, such as from Celsius to Kelvin or Fahrenheit to Rankine. This process is essential for ensuring consistency in thermodynamic calculations and for comparing experimental results across different systems that use different temperature scales. It emphasizes the concept of absolute zero and the relative differences between scales.

Rankine Temperature Scale

The Rankine temperature scale is an absolute temperature scale based on the Fahrenheit degree, similar to the Kelvin scale based on the Celsius degree. In the Rankine scale, 0°R is absolute zero, and each degree increment has the same magnitude as a Fahrenheit degree. This scale is mainly used in engineering applications, particularly in the United States, where Fahrenheit is common.

Gas Constant and Unit Conversion

The gas constant, often symbolized as R, is a fundamental constant in the ideal gas law and thermodynamics, and its value depends on the units used. Converting R to different units, such as from L·atm/(mol·K) to L·atm/(mol·°R), requires understanding the relationship between the temperature scales (Kelvin and Rankine) and ensuring consistent unit usage in calculations involving gases.

van der Waals Equation

The van der Waals equation is a modification of the ideal gas law that accounts for the non-ideal behavior of real gases. It introduces parameters to correct for the molecular volume (b) and intermolecular forces (a), which become significant under conditions of high pressure and low temperature. This equation provides a more accurate description of gas behavior beyond the assumptions of the ideal gas approximation.

Transcript

00:01 This problem essentially is teaching us how we can convert various units, such as the rankin scale and fahrenheit and celsius.

00:12 So for part a of this problem, let's first note that the freezing point of water, h2o, is 32 degrees fahrenheit, and we should know that absolute zero is the exact same thing as negative 273 celsius, which is the same as 0 kelvin or 0r.

00:31 And if we want to know the value of r, we call this a rank and scale.

00:38 It's simply an equation.

00:40 R is equivalent to c or celsius plus 273 multiplied by 9 over 5.

00:49 Now, if we wanted to know the r value of the freezing point of water, we would simply plug in what we know.

00:57 We would have 9 over 5 times 273 because our...

01:02 Degree in celsius is zero, and we'll find that our freezing point in r is 491 .67 degrees in r.

01:14 Now for part b, we have to do a little bit more conversion.

01:19 We're told that we have this value of 0 .0821 degrees celsius per mole times liters atms.

01:28 We want to note this value in r, so we would take 5 over 9 multiplied by 0 .0821 and we would get 0 .056 degrees in r.

01:42 Now on to part c we have a little bit more work to do but we're given a lot of these values.

01:49 We'll be using this equation where we'll have p plus a n squared over v squared times v minus n b which is equivalent to n r t...

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