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Question
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
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Solution 1
Kelvin and Celsius scales are related as:
TC = TK – 273.15 … (i)
Celsius and Fahrenheit scales are related as:
`T_C = 9/5 T_c + 32 ...(ii)
For neon:
TK = 24.57 K
∴TC = 24.57 – 273.15 = –248.58°C
`T_F = 9/5 T_c + 32`
`= 9/5(-248.58) + 32 `
= 415.44ºF
For carbon dioxide:
TK = 216.55 K
∴TC= 216.55 – 273.15 = –56.60°C
`T_F = 9/5(T_C) + 32`
`= 9/5 (-56.60) + 30`
= -69.88ºC
Solution 2
The relation between kelvin scale and Celsius scale is TK – 273.15 =TC => TC=TK– 273.15
For neon `T_K = 24.57K`
`:. T_C = 24.57 - 273.15 = - 248.58 ^@C`
For `CO_2 T_K = 216.55K`
`:.T_C = 216.55 - 273.15 = -56.60^@C`
Also the relation between Kelvin scale and Fahrenheit scale is
`(T_K - 273.15)/100 = (T_F- 32)/180`
`:. T_F = 9/5 (T_K - 273.15) + 32`
Now for neon, `T_K = 24.57 K`
`:.T_F = 9/5 [24.57 - 273.15] + 32 = -415.44 ^@F`
For `CO_2` `T_K = 216.55 K`
`:. T_F = 9/5[216.55 - 273.15] + 32 = -69.88 ^@F`
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