# Conversion confusion: Rankine in ideal gas constant

• Undoubtedly0
In summary: So the conversion should be °R = K*5/9, not K*9/5. In summary, the problem in converting from imperial to SI units is due to a mistake in the temperature conversion. The correct conversion is °R = K*5/9, not K*9/5.
Undoubtedly0
Hi all. I have ran into a problem in converting from imperial to SI units, as follows.

The ideal gas constant for air is often given in imperial units as

$$R = 1716 \frac{ft*lbf}{slug*°R}$$

where

$$1 ft*lbf = 1.356(10^{-3}) kJ$$

$$1 slug = 14.59 kg$$

$$°R = \frac{9}{5}K$$

Thus making these substitutions gives

$$R = 1716 \frac{ft*lbf}{slug*°R} = 1716 \frac{(1.356(10^{-3}) kJ)}{(14.59 kg)(\frac{9}{5}K)} = 0.0886 \frac{kJ}{kg*K}$$

Yet it is quite common knowledge that in reality,

$$R = 0.287 \frac{kJ}{kg*K}$$

Somewhere, then, in my conversion, there must be an error. I found that if I instead reversed the temperature conversion to be °R = K*5/9, the desired result is given - but how could this be? Surely it is must be true that °R = K*9/5.

Thanks for the help.

Last edited:
Undoubtedly0 said:
Hi all. I have ran into a problem in converting from imperial to SI units, as follows.

The ideal gas constant for air is often given in imperial units as

$$R = 1716 \frac{ft*lbf}{slug*°R}$$

where

$$1 ft*lbf = 1.356(10^{-3}) kJ$$

$$1 slug = 14.59 kg$$

$$°R = \frac{9}{5}K$$

Thus making these substitutions gives

$$R = 1716 \frac{ft*lbf}{slug*°R} = 1716 \frac{(1.356(10^{-3}) kJ)}{(14.59 kg)(\frac{9}{5}K)} = 0.0886 \frac{kJ}{kg*K}$$

Yet it is quite common knowledge that in reality,

$$R = 0.287 \frac{kJ}{kg*K}$$

Somewhere, then, in my conversion, there must be an error. I found that if I instead reversed the temperature conversion to be °R = K*5/9, the desired result is given - but how could this be? Surely it is must be true that °R = K*9/5.

Thanks for the help.
$$1\ degree\ R=\frac{5}{9}\ degree\ K$$so there are 5/9 degree K/degree R

## 1. What is the Rankine scale and how does it differ from the Celsius and Fahrenheit scales?

The Rankine scale is a temperature scale that uses absolute zero as its starting point, similar to the Kelvin scale. However, the Rankine scale uses a different unit of measurement - the degree Rankine (°R) - which is equal to one degree Fahrenheit. This means that the Rankine scale has the same intervals as the Fahrenheit scale, but with a different zero point.

## 2. How is the ideal gas constant related to temperature on the Rankine scale?

The ideal gas constant, denoted by the symbol R, is a physical constant that relates the properties of an ideal gas to its temperature, volume, and pressure. On the Rankine scale, the ideal gas constant is given by 1.8 times the value on the Kelvin scale. This is because the Rankine scale has a different unit of temperature (°R) compared to the Kelvin scale (K).

## 3. Can the ideal gas constant be used in calculations involving gases on any temperature scale?

Yes, the ideal gas constant is a universal constant and can be used in calculations involving gases on any temperature scale. However, the value of R will differ depending on the scale being used. It is important to ensure that the units of temperature are consistent when using the ideal gas constant in calculations.

## 4. Why do scientists sometimes use the Rankine scale instead of the Kelvin scale?

The Rankine scale is primarily used in the field of engineering, particularly in the United States. This is because the Fahrenheit scale is still commonly used in many engineering applications in the US, and the Rankine scale has the same intervals as the Fahrenheit scale. It is also sometimes used in thermodynamics calculations as it can make calculations simpler in certain situations.

## 5. Is there a conversion formula for converting between Rankine and other temperature scales?

Yes, there are conversion formulas for converting between Rankine and other temperature scales. For example, to convert from Rankine to Kelvin, you can use the formula K = (°R + 459.67) x 5/9. It is important to note that these conversion formulas will differ depending on the specific temperature scale being converted to or from.

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