Entropy of Ideal Gases

This problem involves comparing the properties of two different isotopes of oxygen, $^{16}\text{O}_2$ and $^{18}\text{O}_2$, when they have the same entropy under specific conditions. We are given that we have 1 mole of each gas in containers of the same volume, and their entropies are equal. We are also instructed to ignore rotational and vibrational contributions to the entropy, focusing only on the translational entropy.

The translational entropy ($S_{trans}$) for an ideal gas is given by a formula that depends on the volume ($V$), the number of moles ($n$), the molar mass ($M$), and the temperature ($T$). For 1 mole of an ideal gas, the translational entropy can be expressed as:

$\qquad S_{trans} = R \ln V + \frac{3}{2} R \ln(M T) + \text{constant}$

where $R$ is the ideal gas constant, and the constant includes terms related to fundamental constants and the number of moles.

We are given that:

• Number of moles of $^{16}\text{O}_2$ ($n_1$) = 1 mole
• Number of moles of $^{18}\text{O}_2$ ($n_2$) = 1 mole
• Volume of container for $^{16}\text{O}_2$ ($V_1$) = Volume of container for $^{18}\text{O}_2$ ($V_2$) = $V$
• Entropy of $^{16}\text{O}_2$ ($S_1$) = Entropy of $^{18}\text{O}_2$ ($S_2$) = $S$
• Temperature of $^{16}\text{O}_2$ ($T_1$) = 300 K

We need to find the temperature of $^{18}\text{O}_2$ ($T_2$).

Since $S_1 = S_2$ and the volume and number of moles are the same, we can write:

$R \ln V + \frac{3}{2} R \ln(M_1 T_1) + \text{constant} = R \ln V + \frac{3}{2} R \ln(M_2 T_2) + \text{constant}$

Subtracting the common terms ($R \ln V$ and the constant) from both sides:

$\frac{3}{2} R \ln(M_1 T_1) = \frac{3}{2} R \ln(M_2 T_2)$

Dividing by $\frac{3}{2} R$:

$\ln(M_1 T_1) = \ln(M_2 T_2)$

Exponentiating both sides:

$M_1 T_1 = M_2 T_2$

This is the key relationship when translational entropy, volume, and number of moles are equal for two different ideal gases.

Calculating Temperature

Now, we need to find the molar masses of $^{16}\text{O}_2$ and $^{18}\text{O}_2$.

• Molar mass of $^{16}\text{O}_2$ ($M_1$) = $2 \times \text{atomic mass of }^{16}\text{O} = 2 \times 16 \, \text{g/mol} = 32 \, \text{g/mol}$
• Molar mass of $^{18}\text{O}_2$ ($M_2$) = $2 \times \text{atomic mass of }^{18}\text{O} = 2 \times 18 \, \text{g/mol} = 36 \, \text{g/mol}$

Using the relationship $M_1 T_1 = M_2 T_2$:

$32 \, \text{g/mol} \times 300 \, \text{K} = 36 \, \text{g/mol} \times T_2$

Solving for $T_2$:

$T_2 = \frac{32 \times 300}{36} \, \text{K}$

$T_2 = \frac{32 \times (12 \times 25)}{12 \times 3} \, \text{K}$

$T_2 = \frac{32 \times 25}{3} \, \text{K}$

$T_2 = \frac{800}{3} \, \text{K}$

$T_2 \approx 266.666... \, \text{K}$

Rounding to two decimal places, the temperature of $^{18}\text{O}_2$ is approximately 266.67 K. Comparing this to the given options, 266.66 is the closest value.

The temperature of $^{18}\text{O}_2$ for the same translational entropy as $^{16}\text{O}_2$ at 300K is approximately 266.66 K.