Heat capacity or thermal capacity is an extensive property of matter, that defines its physical property. Heat Capacity is the amount of heat that must be applied to an object in order to cause a unit change in temperature. Heat capacity is measured in Joules per Kelvin (J/K), which is its SI unit. 

In general, the quantity of heat required to warm a given substance depends on its mass, \(m\), the change in temperature, \(\Delta T\), and the nature of the substance.

The change in temperature of a substance, when a given quantity of heat is absorbed or rejected by it, is characterised by a quantity called the heat capacity of that substance. We define heat capacity, \(S\) of a substance as
\(
S=\frac{\Delta Q}{\Delta T} \dots(11.10)
\)
where \(\Delta Q\) is the amount of heat supplied to the substance to change its temperature from \(T\) to \(T+\Delta T\)

SI Unit of Heat Capacity

• The SI unit for the heat capacity of a substance is Joule per Kelvin ( \(\mathrm{J} / \mathrm{K}\) or \(\mathrm{J} \cdot \mathrm{K}^{-1}\) ). When the temperature is increased by \(1^{\circ} \mathrm{C}\) is the same as an increase of \(1 \mathrm{~K}\), which is the same unit as \(\mathrm{J} /{ }^{\circ} \mathrm{C}\).
• The dimension of Heat capacity is \(\left[\mathrm{L}^2 \mathrm{MT}^{-2} \mathrm{\Theta}^{-1}\right]\).

Specific Heat Capacity

Specific heat capacity is defined as the amount of heat absorbed or given off by a substance to change the temperature of unit mass of it by one unit.

If \(\Delta Q\) stands for the amount of heat absorbed or given off by a substance of mass \(m\) when it undergoes a temperature change \(\Delta T\), then the specific heat capacity, of that substance is given by
\(
s=\frac{S}{m}=\frac{1}{m} \frac{\Delta Q}{\Delta T} \dots(11.11)
\)

If the amount of substance is specified in terms of moles \(\mu\), instead of mass \(m\) in \(\mathrm{kg}\), we can define heat capacity per mole of the substance by
\(
C=\frac{S}{\mu}=\frac{1}{\mu} \frac{\Delta Q}{\Delta T} \dots(11.12)
\)
where \(C\) is known as molar specific heat capacity of the substance. Like \(S, C\) also depends on the nature of the substance and its temperature. The SI unit of molar specific heat capacity is \(\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\).

The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature. The molar heat capacity at constant volume, denoted by \(C_V\), is
\(
C_V=\left(\frac{\Delta Q}{\mu \Delta T}\right)_{\text {constant volume }}
\)
and the molar heat capacity at constant pressure, denoted by \(C_p\), is
\(
C_p=\left(\frac{\Delta Q}{\mu \Delta T}\right)_{\text {constant pressure }}
\)
where \(\mu\) is the amount of the gas in number of moles. Quite often, the term specific heat capacity or specific heat is used for molar heat capacity. It is advised that the unit be carefully noted to determine the actual meaning. The unit of specific heat capacity is \(\mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1}\) whereas that of molar heat capacity is \(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\).

However, in connection with the specific heat capacity of gases, additional conditions may be needed to define \(C\). In this case, heat transfer can be achieved by keeping either pressure or volume constant. If the gas is held under constant pressure during the heat transfer, then it is called the molar specific heat capacity at constant pressure and is denoted by \(C_{\mathrm{p}}\). On the other hand, if the volume of the gas is maintained during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant volume and is denoted by \(C_{\text {v }}\). Table \(11.3\) lists the measured specific heat capacity of some substances at atmospheric pressure and ordinary temperature while Table 11.4 lists the molar specific heat capacities of some gases. From Table \(11.3\) you can note that water has the highest specific heat capacity compared to other substances. For this reason, water is also used as a coolant in automobile radiators, as well as, a heater in hot water bags. Owing to its high specific heat capacity, the water warms up more slowly than land during summer, and consequently wind from the sea has a cooling effect. Now, you can tell why in desert areas, the earth’s surface warms up quickly during the day and cools quickly at night.

Relation Between \(C_p\) and \(C_V\) for an Ideal gas

Consider an amount \(\mu\) (in moles) of an ideal gas kept in a rigid container at an initial pressure \(p\), volume \(V\) and temperature \(T\). An amount \((d Q)_V\) of heat is supplied to the gas. Its temperature rises from \(T\) to \(T+d T\) whereas the volume remains constant. The work done by the gas is \(d W=0\). From the first law of thermodynamics, the internal energy changes by \(d U\) such that
\(
(d Q)_V=d U \dots(i)
\)
Now suppose, the same sample of the gas is taken in a vessel with a movable light piston of cross-sectional area \(A\) (figure below). The initial pressure, volume and temperature are \(p, V\) and \(T\). The piston is pushed by a constant external force \(F=p A\). The pressure inside the gas thus remains constant at the value \(p\).

The gas is given heat \((d Q)_p\) which raises the temperature by the same amount \(d T\). The piston moves out so that the volume increases by \(d V\). The work done by the gas is \(p d V\) and from the first law of thermodynamics,
\(
(d Q)_p=d U^{\prime}+p d V \dots(ii)
\)
For an ideal gas, \(p V=\mu R T\).
As the pressure remains constant,
and hence,
\(
p(V+d V)=\mu R(T+d T)
\)
\(p d V=\mu R d T\).
From (ii),
\(
(d Q)_p=d U^{\prime}+\mu R d T \text {. } \dots(iii)
\)
As the temperature rises by the same amount in the two cases and the internal energy of an ideal gas depends only on its temperature,
\(
d U=d U^{\prime} .
\)
Thus, from (i) and (iii),
\(
\begin{aligned}
(d Q)_p & =(d Q)_V+\mu R d T \\
\text { or, } \quad \frac{1}{\mu} \frac{(d Q)_p}{d T} & =\frac{1}{\mu} \frac{(d Q)_V}{d T}+R \dots(iv)
\end{aligned}
\)
But \((d Q)_p\) is the heat given to increase the temperature of the gas by \(d T\) at constant pressure. Thus, by definition,
\(
C_p=\frac{1}{\mu} \frac{(d Q)_p}{d T}
\)
\(
\text { Similarly, } \quad C_V=\frac{1}{\mu} \frac{(d Q)_V}{d T}
\)
and (iv) becomes
\(
\begin{aligned}
C_p & =C_V+R \\
\text { or, } \quad C_p-C_V & =R .
\end{aligned}
\)
Relations of \(C_V\) with energy
From (i), or, or,
\(
\begin{aligned}
(d Q)_V & =d U \\
\frac{1}{\mu} \frac{(d Q)_V}{d T} & =\frac{1}{\mu} \frac{d U}{d T} \\
C_V & =\frac{1}{\mu} \frac{d U}{d T} \\
d U & =\mu C_V d T .
\end{aligned}
\)
\(
\begin{aligned}
& \text { or, } d U=\mu C_V d T . \\
& \text { Taking the energy to be zero at } T=0,
\end{aligned}
\)
\(
U=\mu C_V T \text {. }
\)

Note: Some books use \(n\) instead of \(\mu\). 

Example 1: A \(30.5 \mathrm{~g}\) sample of an alloy at \(93.0^{\circ} \mathrm{C}\) is placed into \(50.0 \mathrm{~g}\) of water at \(22.0^{\circ} \mathrm{C}\) in an insulated coffee cup with a heat capacity of \(9.2 \mathrm{~J} / \mathrm{K}\). If the final temperature of the system is \(31.1^{\circ} \mathrm{C}\), what is the specific heat capacity of the alloy?

Solution:

Heat absorbed \(=\) heat lost
then the specific heat capacity, of that substance is given by
\(
s=(1 / m)(\Delta Q / \Delta T)
\)
Rearrange the above expression,
\(
\begin{aligned}
& \Delta Q=s m \Delta T \\
& \Delta Q_{\text {alloy }}=\Delta Q_{\text {water }}+\Delta Q_{c u p}
\end{aligned}
\)
The temperature of the water is equal to the temperature of \(22.0^{\circ} \mathrm{C}\)
The temperature of the alloy is \(93.0^{\circ} \mathrm{C}\).
The final temperature is \(31.1^{\circ} \mathrm{C}\).
\(
\begin{aligned}
& 30.5 \times(93.0-31.1) s=9.2 \times(31.1-22.0)+50.0 \times 4.2 \times(31.1-22.0) \\
& 1887.95 s=1994.72 \\
& s=1.057 \mathrm{~J} / \mathrm{gK}
\end{aligned}
\)

Example 2: The specific heat of water is \(4.18 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)\). Calculate the molar heat capacity of water. Express your answer to three significant figures and include the appropriate units.

Solution:

The specific heat of water is \(4.18 \mathrm{~J} /\left(g^{\circ} \mathrm{C}\right)\).
The expression to convert gram into mole is
\(4.18 \mathrm{~J} / \mathrm{gC} \times(18.0 \mathrm{~g} / \mathrm{mole})=75.24 \mathrm{~J} / \mathrm{mole} C\)
Hence, the molar heat capacity of water is \(75.24 \mathrm{~J} /\) mole \(\mathrm{C}\)

Example 3: A 88.3 g sample of metal at \(95.24 \mathrm{C}\) is added to \(35.10 \mathrm{~g}\) of water that is initially at \(17.27^{\circ} \mathrm{C}\). The final temperature of both the water and the metal is \(29.20^{\circ} \mathrm{C}\). The specific heat of water is \(4.184 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)\). Calculate the specific heat of the metal.

Solution:

Given,
Mass of metal is \(88.3 \mathrm{~g}\)
The initial temperature of the metal is \(95.24^{\circ} \mathrm{C}\).
Mass of water: \(35.10 \mathrm{~g}\).
The initial temperature of the water is \(17.27^{\circ} \mathrm{C}\).
The final temperature of the water and the metal is \(29.20^{\circ} \mathrm{C}\).
The specific heat of water is \(4.184 \mathrm{~J} /\left(g^{\circ} \mathrm{C}\right)\).
Therefore, the expression where the energy from the hotter metal transfers to the cooler water is
\(
-m_o C_o \Delta T_o=m_w C_w \Delta T_w
\)
Where
\(m_0=\) mass of a metal object
\(\Delta T_o=\) temperature change of metal object
\(C_0=\) specific heat capacity of metal object
\(m_w=\) mass of water
\(\Delta T_w=\) temperature change of water
\(C_w=\) specific heat capacity of water
Rearrange the above expression,
\(
C_o=\left(m_w C_W \Delta T_w\right) /\left(m_o \Delta T_o\right)
\)
Substitute the values in the above expression,
\(
\begin{aligned}
& C_o=[35.104 .184(29.20-17.27)] /[88.3(29.20-95.24)] \\
& =0.301 \mathrm{~J} / g^{\circ} \mathrm{C}
\end{aligned}
\)

Example 4: \(0.32 \mathrm{~g}\) of oxygen is kept in a rigid container and is heated. Find the amount of heat needed to raise the temperature from \(25^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\). The molar heat capacity of oxygen at constant volume is \(20 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).

Solution:

The molecular weight of oxygen \(=32 \mathrm{~g} \mathrm{~mol}^{-1}\).
The amount of the gas in moles is
\(
n=\frac{0.32 \mathrm{~g}}{32 \mathrm{~g} \mathrm{~mol}}=0.01 \mathrm{~mol} \text {. }
\)
The amount of heat needed is \(Q=n C_V \Delta T\)
\(
=(0.01 \mathrm{~mol})\left(20 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(10 \mathrm{~K})=2.0 \mathrm{~J} .
\)

Example 5: A tank of volume \(0.2 \mathrm{~m}^3\) contains helium gas at a temperature of \(300 \mathrm{~K}\) and pressure \(1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\). Find the amount of heat required to raise the temperature to \(400 \mathrm{~K}\). The molar heat capacity of helium at constant volume is \(3.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\). Neglect any expansion in the volume of the tank.

Solution:

The amount of the gas in moles is
\(
\begin{aligned}
n & =\frac{p V}{R T} \\
& =\frac{\left(1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}\right)\left(0.2 \mathrm{~m}^3\right)}{\left(8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})}=8.0 \mathrm{~mol} .
\end{aligned}
\)
The amount of heat required is
\(
\begin{aligned}
\Delta Q & =n C_V \Delta T \\
& =(8.0 \mathrm{~mol})\left(3.0 \mathrm{cal} \mathrm{mol}{ }^{-1} \mathrm{~K}^{-1}\right)(100 \mathrm{~K})=2400 \mathrm{cal} .
\end{aligned}
\)

Example 6: The molar heat capacity of a gas at constant volume is found to be \(5 \mathrm{cal} \mathrm{mol} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\). Find the ratio \(\gamma=C_p / C_V\) for the gas. The gas constant \(R=2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\).

Solution:

We have \(C_V=5 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\).
Thus, \(C_p=C_{\mathrm{V}}+R=5 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}+2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\) \(=7 \mathrm{cal} \mathrm{mol}{ }^{-1} \mathrm{~K}^{-1}\)
or, \(\quad \frac{C_p}{C_V}=\frac{7}{5}=1.4\).

Determination of \(C_v\) Of a Gas

Suppose, the mass of the gas taken \(\quad=m_1\)
the mass of the extra steam condensed \(=m_2\)
Initial temperature of the gas \(\quad=\theta_1\)
Final temperature of the gas \(\quad=\theta_2\)
Specific latent heat of vaporization of water \(=L\).
If the molecular weight of the gas is \(M\), the amount of the gas in moles is \(n=m_1 / M\). The heat lost by the steam is \(=m_2 L\) and the heat gained by the gas is \(n C_V\left(\theta_2-\theta_1\right)\).
Thus, \(m_2 L=n C_V\left(\theta_2-\theta_1\right)=\frac{m_1}{M} C_V\left(\theta_2-\theta_1\right)\)
or, \(\quad C_V=\frac{M m_2 L}{m_1\left(\theta_2-\theta_1\right)}\).

Determination of \(C_p\) Of a Gas

Let the water equivalent of the calorimeter together with the coil \(D \quad=W\)
mass of the water \(\quad=m\)
temperature of the oil bath \(=\theta_1\)
initial temperature of water \(=\theta_2\)
final temperature of water \(=\theta_3\)
and the amount of the gas (in moles) passed through the water \(=n\).

The gas at temperature \(\theta_1\) enters the coil \(D\). At the beginning of the experiment, the gas leaves the coil \(D\) at temperature \(\theta_2\). This temperature gradually increases and at the end of the experiment, it becomes \(\theta_3\). The average temperature of the gas leaving the coil \(D\) is, therefore, \(\frac{\theta_2+\theta_3}{2}\). The heat lost by the gas is
\(
\Delta Q=n C_p\left[\theta_1-\frac{\theta_2+\theta_3}{2}\right] \text {. } \dots(i)
\)
This heat is used to increase the temperature of the calorimeter, the water, and the coil \(D\) from \(\theta_2\) to \(\theta_3\). The heat received by them is
\(
\Delta Q=(W+m) s\left(\theta_3-\theta_2\right) \dots(ii)
\)
where \(s\) is the specific heat capacity of water. From (i) and (ii),
\(
n C_p\left[\theta_1-\frac{\theta_2+\theta_3}{2}\right]=(W+m) s\left(\theta_9-\theta_2\right)
\)
\(
\text { or, } \quad C_p=\frac{(W+m) s\left(\theta_3-\theta_2\right)}{n\left[\theta_1-\frac{\theta_2+\theta_3}{2}\right]} \text {. } \dots(iii)
\)

Determination of \(n\)

Suppose the difference in the mercury levels in the manometer \(M\) is \(h\) and the atmospheric pressure is equal to a height \(H\) of mercury. The pressure in the tank is \(p\), equal to a height \(H+h\) of mercury. By noting the difference \(h\) at the beginning and at the end of the experiment, the initial pressure \(p_1\) and the final pressure \(p_2\) are determined. Assuming the gas to be ideal,
and
\(
\begin{aligned}
& p_1 V=n_1 R T \\
& p_2 V=n_2 R T .
\end{aligned}
\)
Here \(n_1, n_2\) are the amounts of the gas (in moles) in the tank at the beginning and at the end respectively, \(V\) is the volume of the tank and \(T\) is the absolute temperature of the tank. Thus,

\(
\begin{aligned}
&\left(p_1-p_2\right) V=\left(n_1-n_2\right) R T \\
& \text { or, } \quad n=n_1-n_2=\frac{\left(p_1-p_2\right) V}{R T}
\end{aligned}
\)

Isothermal Process

A process in a system is called isothermal if the temperature of the system remains constant. In the case of an ideal gas, the internal energy remains constant in such a process. The amount of heat supplied is equal to the work done by the gas. The gas obeys Boyle’slaw and the work done by the gas is

\(
W=\int_{V_i}^{V_f} p d V=\int_{V_i}^{V_f} n \frac{R T}{V} d V=n R T \ln \left(V_f / V_i\right)
\)

This is also the amount of heat given as the volume of the gas is changed from \(V_i\) to \(V_f\). As the change in temperature is zero, the molar heat capacity in such a process is
\(
C_{\text {isothermal }}=\frac{\Delta Q}{n \Delta T}=\text { infinity. }
\)

Adiabatic Process

A process on a system is called adiabatic if no heat is supplied to it or extracted from it. In such a case, the temperature changes without adding any heat. The molar heat capacity in such a process is
\(
C_{\text {adiabatic }}=\frac{\Delta Q}{n \Delta T}=\text { zero. }
\)
The work done by the gas in an adiabatic process equals the decrease in its internal energy. Thus, if a gas enclosed in a container with adiabatic walls expands, the internal energy decreases and hence the temperature falls. If the gas is compressed adiabatically, the temperature rises.

Relations Between \(p, V, T\) in a Reversible Adiabatic Process

Consider an adiabatic process on an ideal gas. During a short part of the process, the pressure, the volume and the temperature change from \(p, V, T\) to \(p+d p, V+d V\) and \(T+d T\) respectively. The internal energy changes from \(U\) to \(U+d U\). As the amount of heat supplied is zero, the first law of thermodynamics gives
\(
0=d U+p d V \dots(i)
\)
We have, \(\quad C_V=\frac{1}{n} \frac{d U}{d T}\)
or, \(\quad d U=n C_v d T\).
Thus, from (i),
\(
n C_V d T+p d V=0 \text {. } \dots(ii)
\)
As the gas is ideal,
\(
p V=n R T
\)
or, \(\quad p d V+V d p=n R d T\)
or, \(\quad d T=\frac{p d V+V d p}{n R}\).
Substituting this expression for \(d T\) in (ii),
\(C_V\left(\frac{p d V+V d p}{R}\right)+p d V=0\)
or, \(\quad\left(C_V+R\right) p d V+C_V V d p=0\)
or, \(\quad C_p p d V+C_V V d p=0\)
or, \(\quad \gamma \frac{d V}{V}=-\frac{d p}{p}\)
where \(\gamma=C_p / C_V\).
Let the initial pressure and volume be \(p_i\) and \(V_i\) respectively and the final pressure and volume be \(p_f\) and \(V_f\) respectively. Then
\(
\begin{aligned}
& \int_{v_i}^{v_f} \gamma \frac{d V}{V}=-\int_{p_i}^{p_f} \frac{d p}{p} . \\
& \text { or, } \quad \gamma \ln \frac{V_f}{V_i}=-\ln \frac{p_f}{p_i} \\
& \text { or, } \quad \ln \left(\frac{V_f}{V_i}\right)^\gamma=\ln \left(\frac{p_i}{p_f}\right) \\
& \text { or, } \quad \frac{V_f^\gamma}{V_i^\gamma}=\frac{p_i}{p_f} \\
& \text { or, } \quad p_i V_i^\gamma=p_f V_f^\gamma \\
& \text { or, } \quad p V^\gamma=\text { constant. } \dots(iii)\\
\end{aligned}
\)
Thus, \(p V^\gamma\) remains constant in a reversible adiabatic process.
Relation between \(p\) and \(T\)
We have \(\quad p V=n R T\)
or, \(\quad V=\frac{n R T}{p}\).
Putting in (iii),
\(
\begin{aligned}
& p\left(\frac{n R T}{p}\right)^\gamma=\text { constant } \\
& \text { or, } \quad p^{1-\gamma} T^\gamma=\text { constant }
\end{aligned}
\)
or, \(\quad \frac{T^\gamma}{p^{\gamma-1}}=\text { constant } \dots(iv)\)
Relation between \(V\) and \(T\)
We have \(\quad p V=n R T\)
or, \(\quad p=\frac{n R T}{V}\).
Putting in (iii),
\(
\left(\frac{n R T}{V}\right) V^\gamma=\text { constant }
\)
or,
\(T V^{\gamma-1}=\text { constant } \dots(v)\)

Example 7: Dry air at \(15^{\circ} \mathrm{C}\) and 10 atm is suddenly released at atmospheric pressure. Find the final temperature of the \(\operatorname{air}\left[C_p / C_V=1 \cdot 41\right]\).

Solution:

As the air is suddenly released, it does not get time to exchange heat with the surrounding. Thus the process is adiabatic. Assuming the process to be reversible,
\(
\frac{T^\gamma}{p^{\gamma-1}}=\text { constant }
\)
or,
\(\frac{T_1^\gamma}{p_1^{\gamma-1}}=\frac{T_2^\gamma}{p_2^{\gamma-1}}\)
\(\left(\frac{T_2}{T_1}\right)^\gamma=\left(\frac{p_2}{p_1}\right)^{\gamma-1}\)
\(T_2=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}\).
Taking \(p_1=10 \mathrm{~atm}, p_2=1\) atm, \(\gamma=1.41\) and \(T_1=(273+15) \mathrm{K}=288 \mathrm{~K}\), the final temperature is \(T_2=148 \mathrm{~K}\).

Work Done In An Adiabatic Process

Suppose a sample of gas has initial pressure \(p_1\) and initial volume \(V_1\). In an adiabatic process, the pressure and volume change to \(p_2\) and \(V_2\). We have
\(
p V^\gamma=p_1 V_1^\gamma=p_2 V_2^\gamma=K \text {. } \dots(i)
\)
Thus, \(\quad p=\frac{K}{V^\gamma}\).
The work done by the gas in the process is
\(
\begin{aligned}
W & =\int_{V_1}^{V_2} p d V=\int_{V_1}^{V_2} \frac{K}{V^\gamma} d V \\
& =\frac{1}{1-\gamma}\left[\frac{K}{V_2^{\gamma-1}}-\frac{K}{V_1^{\gamma-1}}\right] .
\end{aligned}
\)
From (i), \(\frac{K}{V_2^\gamma}=p_2\) and \(\frac{K}{V_1^\gamma}=p_1\).
Thus,
\(
\begin{aligned}
W & =-\frac{1}{\gamma-1}\left(p_2 V_2-p_1 V_1\right) \\
= & \frac{p_1 V_1-p_2 V_2}{\gamma-1}
\end{aligned}
\)

Example 8: \(\text { Calculate the internal energy of } 1 \mathrm{~g} \text { of oxygen at STP. }\)

Solution:

Oxygen is a diatomic gas. The average energy per molecule is, therefore, \(\frac{5}{2} k T\) and the average energy per mole is \(\frac{5}{2} R T\). As the molecular weight of oxygen is \(32 \mathrm{~g} \mathrm{~mol}^{-1}, 1 \mathrm{~g}\) of oxygen has
\(
n=\frac{1 \mathrm{~g}}{32 \mathrm{~g} \mathrm{~mol}^{-1}}=\frac{1}{32} \mathrm{~mol} .
\)
The temperature of oxygen is \(273 \mathrm{~K}\). Thus, the internal energy is
\(
\begin{aligned}
U & =n\left(\frac{5}{2} R T\right) \\
& =\left(\frac{1}{32} \mathrm{~mol}\right)\left(\frac{5}{2}\right)\left(8 \cdot 31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(273 \mathrm{~K}) = 177 J
\end{aligned}
\)