Kvant Physics Problem 4

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Verdicts: PASS + PASS
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Problem

A cylinder contains a purified gas, but it is unknown which one. To raise the temperature of 1 kg of this gas by one degree at constant pressure requires 958.4 J, and at constant volume, 704.6 J. What gas is it?

III All-Union Physics Olympiad

Setup and Assumptions

A sample of an unknown purified gas is contained in a cylinder. The measured specific heat capacity at constant pressure is

$$c_p = 958.4\ \frac{\text{J}}{\text{kg}\cdot\text{K}},$$

and the specific heat capacity at constant volume is

$$c_v = 704.6\ \frac{\text{J}}{\text{kg}\cdot\text{K}}.$$

The task is to identify the gas.

The gas is assumed to behave as an ideal gas. The measured heat capacities are taken to be constant over the temperature interval considered. Effects of intermolecular forces, dissociation, ionization, and thermal excitation of additional degrees of freedom are neglected.

Physical Principles

For an ideal gas, the difference between the specific heat capacities is related to the specific gas constant:

$$c_p - c_v = R_s,$$

where $R_s$ is the gas constant per unit mass.

The specific gas constant is connected with the universal gas constant $R$ and the molar mass $M$ by

$$R_s = \frac{R}{M}.$$

For an ideal gas, the ratio of heat capacities is

$$\gamma = \frac{c_p}{c_v}.$$

For a monatomic ideal gas,

$$\gamma = \frac{5}{3}.$$

For a diatomic ideal gas with rotational degrees of freedom excited and vibrational degrees of freedom neglected,

$$\gamma = \frac{7}{5}.$$

These characteristic values can be used to identify the molecular structure of the gas.

Derivation

From the relation between the heat capacities,

$$R_s = c_p - c_v.$$

Substituting the given values,

$$R_s = 958.4 - 704.6 = 253.8\ \frac{\text{J}}{\text{kg}\cdot\text{K}}.$$

Using

$$R_s = \frac{R}{M},$$

the molar mass is

$$M = \frac{R}{R_s}.$$

With

$$R = 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}},$$

we obtain

$$M = \frac{8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}}} {253.8\ \frac{\text{J}}{\text{kg}\cdot\text{K}}} = 0.03276\ \frac{\text{kg}}{\text{mol}}.$$

Expressing the result in grams per mole,

$$M = 32.76\ \frac{\text{g}}{\text{mol}}.$$

Next, calculate the heat-capacity ratio:

$$\gamma = \frac{c_p}{c_v} = \frac{958.4}{704.6} = 1.360.$$

This value is close to the theoretical value

$$\gamma = 1.4$$

for a diatomic gas.

Among common diatomic gases, the molar mass closest to

$$32.76\ \frac{\text{g}}{\text{mol}}$$

is molecular oxygen, whose molar mass is

$$32.0\ \frac{\text{g}}{\text{mol}}.$$

A consistency check can be made using the known specific gas constant of oxygen:

$$R_s=\frac{8.314}{0.032} \approx 259.8\ \frac{\text{J}}{\text{kg}\cdot\text{K}},$$

which is very close to the measured value

$$253.8\ \frac{\text{J}}{\text{kg}\cdot\text{K}}.$$

The small discrepancy is consistent with experimental uncertainty and the ideal-gas approximation.

Result

The symbolic expression for the molar mass is

$$M=\frac{R}{c_p-c_v}.$$

Substituting the data,

$$M = \frac{8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}}} {(958.4-704.6)\ \frac{\text{J}}{\text{kg}\cdot\text{K}}} = 0.03276\ \frac{\text{kg}}{\text{mol}} = 32.76\ \frac{\text{g}}{\text{mol}}.$$

The measured ratio

$$\gamma=\frac{958.4}{704.6}=1.360$$

indicates a diatomic gas.

Thus the gas is

$$\boxed{\text{oxygen }(\mathrm{O_2})}.$$

Sanity Checks

The quantity $c_p-c_v$ has units

$$\frac{\text{J}}{\text{kg}\cdot\text{K}},$$

which are the units of a specific gas constant, so the equation

$$R_s=c_p-c_v$$

is dimensionally correct.

The calculated molar mass,

$$M\approx 33\ \frac{\text{g}}{\text{mol}},$$

lies in the range of common diatomic gases. Nitrogen has molar mass $28\ \text{g/mol}$, oxygen has $32\ \text{g/mol}$, and chlorine has about $71\ \text{g/mol}$. The value is closest to oxygen.

The heat-capacity ratio satisfies

$$1<\gamma<\frac{5}{3},$$

as required for a gas with internal degrees of freedom beyond pure translation. Its proximity to $1.4$ supports the identification as a diatomic gas.

The step most vulnerable to error is the use of

$$R_s=c_p-c_v.$$

Confusing the specific gas constant $R_s$ with the universal gas constant $R$ would produce an incorrect molar mass by a factor of approximately $10^3$.