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Chapter 1: Problem 24

The universal gas constant \(R_{0}\) is equal to \(49,700 \mathrm{ft}^{2} /\left(\mathrm{s}^{2} \cdot^{\circ} \mathrm{R}\right)\) or \(8310 \mathrm{m}^{2} /\left(\mathrm{s}^{2} \cdot \mathrm{K}\right) .\) Show that these two magnitudes are equal.

Short Answer

Expert verified
After converting the units and simplifying, we find that both magnitudes are equal. Thus, \(49,700 \mathrm{ft}^{2} / \mathrm{s}^{2} °\mathrm{R} = 8310 \mathrm{m}^{2} / \mathrm{s}^{2}K\)

Step by step solution

01

Conversion of feet to meters

Firstly, convert feet to meters: Knowing that 1ft = 0.3048m, the conversion can be expressed mathematically as \(49,700 (ft)^{2} / s^{2}°R = 49,700 (0.3048m/1ft)^{2} / s^{2}°R\)
02

Conversion of Fahrenheit to Kelvin

Secondly, convert Fahrenheit to Kelvin: The conversion factor between Fahrenheit and Kelvin is \(K = (°F - 32) × 5/9 + 273.15\). Thus the Rankine temperature scale (°R) used in the English system, where 0 °R = absolute zero and uses Fahrenheit-degree intervals, can be converted to Kelvin (K) in the metric system by knowing that 1 °R = 0.55556 K. Therefore, the conversion can be expressed as \(49,700 (0.3048m/1ft)^{2} / s^{2}°R = 49,700 (0.3048m/1ft)^{2} / s^{2}(1K / 0.55556°R)\)
03

Evaluation of the expression

Simplify the above expression to obtain the resulting magnitude in Metric units.

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Most popular questions from this chapter

Develop a computer progrem for calculating the density of an ideal gas when the gas pressure in pascals (abs), the temperature in degrees Celsius, and the gas constant in \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) are specified. Plot the density of helium as a function of temperature from \(0^{\circ} \mathrm{C}\) to \(200^{\circ} \mathrm{C}\) and pressures of \(50,100,150,\) and \(200 \mathrm{kPa}(\mathrm{abs})\).

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