# 2.3 Heat capacity and equipartition of energy

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By the end of this section, you will be able to:
• Solve problems involving heat transfer to and from ideal monatomic gases whose volumes are held constant
• Solve similar problems for non-monatomic ideal gases based on the number of degrees of freedom of a molecule
• Estimate the heat capacities of metals using a model based on degrees of freedom

In the chapter on temperature and heat, we defined the specific heat capacity with the equation $Q=mc\text{Δ}T,$ or $c=\left(1\text{/}m\right)Q\text{/}\text{Δ}T$ . However, the properties of an ideal gas depend directly on the number of moles in a sample, so here we define specific heat capacity in terms of the number of moles, not the mass. Furthermore, when talking about solids and liquids, we ignored any changes in volume and pressure with changes in temperature—a good approximation for solids and liquids, but for gases, we have to make some condition on volume or pressure changes. Here, we focus on the heat capacity with the volume held constant. We can calculate it for an ideal gas.

## Heat capacity of an ideal monatomic gas at constant volume

We define the molar heat capacity at constant volume ${C}_{V}$ as

${C}_{V}=\frac{1}{n}\phantom{\rule{0.2em}{0ex}}\frac{Q}{\text{Δ}T},\text{with}\phantom{\rule{0.2em}{0ex}}V\phantom{\rule{0.2em}{0ex}}\text{held constant}.$

This is often expressed in the form

$Q=n{C}_{V}\text{Δ}T.$

If the volume does not change, there is no overall displacement, so no work is done, and the only change in internal energy is due to the heat flow $\text{Δ}{E}_{\text{int}}=Q.$ (This statement is discussed further in the next chapter.) We use the equation ${E}_{\text{int}}=3nRT\text{/}2$ to write $\text{Δ}{E}_{\text{int}}=3nR\text{Δ}T\text{/}2$ and substitute $\text{Δ}E$ for Q to find $Q=3nR\text{Δ}T\text{/}2$ , which gives the following simple result for an ideal monatomic gas:

${C}_{V}=\frac{3}{2}R.$

It is independent of temperature, which justifies our use of finite differences instead of a derivative. This formula agrees well with experimental results.

In the next chapter we discuss the molar specific heat at constant pressure ${C}_{p},$ which is always greater than ${C}_{V}.$

## Calculating temperature

A sample of 0.125 kg of xenon is contained in a rigid metal cylinder, big enough that the xenon can be modeled as an ideal gas, at a temperature of $20.0\phantom{\rule{0.2em}{0ex}}\text{°C}$ . The cylinder is moved outside on a hot summer day. As the xenon comes into equilibrium by reaching the temperature of its surroundings, 180 J of heat are conducted to it through the cylinder walls. What is the equilibrium temperature? Ignore the expansion of the metal cylinder.

## Solution

1. Identify the knowns: We know the initial temperature ${T}_{1}$ is $20.0\phantom{\rule{0.2em}{0ex}}\text{°C}$ , the heat Q is 180 J, and the mass m of the xenon is 0.125 kg.
2. Identify the unknown. We need the final temperature, so we’ll need $\text{Δ}T$ .
3. Determine which equations are needed. Because xenon gas is monatomic, we can use $Q=3nR\text{Δ}T\text{/}2.$ Then we need the number of moles, $n=m\text{/}M.$
4. Substitute the known values into the equations and solve for the unknowns.
The molar mass of xenon is 131.3 g, so we obtain
$n=\frac{125\phantom{\rule{0.2em}{0ex}}\text{g}}{131.3\phantom{\rule{0.2em}{0ex}}\text{g/mol}}=0.952\phantom{\rule{0.2em}{0ex}}\text{mol,}$

$\text{Δ}T=\frac{2Q}{3nR}=\frac{2\left(180\phantom{\rule{0.2em}{0ex}}\text{J}\right)}{3\left(0.952\phantom{\rule{0.2em}{0ex}}\text{mol}\right)\left(8.31\phantom{\rule{0.2em}{0ex}}\text{J/mol}·\text{°}\text{C}\right)}=15.2\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$

Therefore, the final temperature is $35.2\phantom{\rule{0.2em}{0ex}}\text{°C}$ . The problem could equally well be solved in kelvin; as a kelvin is the same size as a degree Celsius of temperature change, you would get $\text{Δ}T=15.2\phantom{\rule{0.2em}{0ex}}\text{K}\text{.}$

## Significance

The heating of an ideal or almost ideal gas at constant volume is important in car engines and many other practical systems.

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Q=mcΔT
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where The letter "Q" is the heat transferred in an exchange in calories, "m" is the mass of the substance being heated in grams, "c" is its specific heat capacity and the static value, and "ΔT" is its change in temperature in degrees Celsius to reflect the change in temperature.
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15 degree Celsius
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15 degree
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ok I think is just conversion
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15 degree Celsius to Fahrenheit
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0 degree Celsius = 32 Fahrenheit
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15 degree Celsius = (15×1.8)+32 =59 Fahrenheit
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I dont understand
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the question said you should convert 15 degree Celsius to Fahrenheit
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what is d final ans for Fahrenheit and Celsius
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it said what is temperature change in Fahrenheit and Celsius
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the 15 is already in Celsius
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So the final answer for Fahrenheit is 59
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