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Chapter 8: Problem 1085

What is the value of absolute temperature on the Celsius Scale? (A) \(-273.15^{\circ} \mathrm{C}\) (B) \(100^{\circ} \mathrm{C}\) (C) \(-32^{\circ} \mathrm{C}\) (D) \(0^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The value of absolute temperature on the Celsius Scale is \(-273.15 ^{\circ} \mathrm{C}\).

Step by step solution

01

Relationship between Celsius and Kelvin scales

The relationship between the Celsius (°C) and Kelvin (K) temperature scales is given by the following formula: \(T_{K} = T_{°C} + 273.15\) Where \(T_{K}\) is the temperature in Kelvin and \(T_{°C}\) is the temperature in Celsius.
02

Define absolute zero

Absolute zero is the lowest possible temperature where atoms would stop moving. This temperature is defined as \(0 \, K\). We need to find the equivalent temperature in Celsius to find the absolute temperature on the Celsius Scale.
03

Calculate the equivalent temperature in Celsius

Using the relationship between Celsius and Kelvin, we plug in the value for absolute zero on the Kelvin scale and solve for Celsius: \(0 \, K = T_{°C} + 273.15\) Now, solve for \(T_{°C}\): \(T_{°C} = 0 \, K - 273.15\) \(T_{°C} = -273.15 ^{\circ} \mathrm{C}\)
04

Choose the correct answer

The value of absolute zero on the Celsius Scale is \(-273.15 ^{\circ} \mathrm{C}\). Hence, the correct answer is: (A) \(-273.15^{\circ} \mathrm{C}\)

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Most popular questions from this chapter

\(\mathrm{Cp}\) and Cv denote the specific heat of oxygen per unit mass at constant Pressure and volume respectively, then (A) \(\mathrm{cp}-\mathrm{cv}=(\mathrm{R} / 16)\) (B) \(\mathrm{Cp}-\mathrm{Cv}=\mathrm{R}\) (C) \(\mathrm{Cp}-\mathrm{Cv}=32 \mathrm{R}\) (D) \(\mathrm{Cp}-\mathrm{Cv}=(\mathrm{R} / 32)\)

What is the relationship Pressure and temperature for an ideal gas undergoing adiabatic Change. (A) \(\mathrm{PT}^{\gamma}=\) Const (B) \(\mathrm{PT}^{-1+\gamma}=\) Const (C) \(\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma}=\) Const (D) \(\mathrm{P}^{\gamma-1} \mathrm{~T}^{\gamma}=\) Const

Each molecule of a gas has \(\mathrm{f}\) degrees of freedom. The radio \(\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma\) for the gas is (A) \(1+(\mathrm{f} / 2)\) (B) \(1+(1 / \mathrm{f})\) (C) \(1+(2 / \mathrm{f})\) (D) \(1+\\{(\mathrm{f}-1) / 3\\}\)

A thermodynamic Process in which temperature \(\mathrm{T}\) of the system remains constant throughout Variable \(\mathrm{P}\) and \(\mathrm{V}\) may Change is called (A) Isothermal Process (B) Isochoric Process (C) Isobasic Process (D) None of this

For an isothermal expansion of a Perfect gas, the value of $(\Delta \mathrm{P} / \mathrm{P})$ is equal to (A) \(-\gamma^{(1 / 2)}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (B) \(-\gamma\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (C) \(-\gamma^{2}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) \((\mathrm{D})-(\Delta \mathrm{V} / \mathrm{V})\)
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