Question

Hydrogen sulfide, H2S, has the characteristic odor of rotten eggs. If a sample of H2S gas at 3.54 atm and 28.8 °C in a 15.4 ml container is allowed to enter a new container at 9.13 atm and 59.9 °C, what is the volume of the new container? Be sure your answer has the correct number of significant figures.

Answer 1

To find the volume of the new container, we can use the combined gas law, which relates pressure, volume, and temperature of a gas. The combined gas law is expressed as:

\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]

where:
- \(P_1\) and \(P_2\) are the initial and final pressures,
- \(V_1\) and \(V_2\) are the initial and final volumes,
- \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.

First, we need to convert the temperatures from Celsius to Kelvin:

\[
T_1 = 28.8 \, °C + 273.15 = 302.95 \, K
\]
\[
T_2 = 59.9 \, °C + 273.15 = 333.05 \, K
\]

Next, we can substitute the known values into the combined gas law formula. We know:
- \(P_1 = 3.54 \, \text{atm}\)
- \(V_1 = 15.4 \, \text{mL}\)
- \(T_1 = 302.95 \, K\)
- \(P_2 = 9.13 \, \text{atm}\)
- \(T_2 = 333.05 \, K\)

We are solving for \(V_2\).

Setting up the equation:

\[
\frac{(3.54 \, \text{atm})(15.4 \, \text{mL})}{302.95 \, K} = \frac{(9.13 \, \text{atm})(V_2)}{333.05 \, K}
\]

Now, we can solve for \(V_2\):

\[
V_2 = \frac{(3.54 \, \text{atm})(15.4 \, \text{mL})(333.05 \, K)}{(9.13 \, \text{atm})(302.95 \, K)}
\]

Calculating the numerator:

\[
3.54 \times 15.4 \times 333.05 = 17379.0987
\]

Calculating the denominator:

\[
9.13 \times 302.95 = 2767.18
\]

Now we can divide the two results:

\[
V_2 = \frac{17379.0987}{2767.18} \approx 6.284 \, \text{mL}
\]

Finally, we must consider significant figures. The least number of significant figures in the given values is 3 (from 15.4 mL). Thus, we round \(V_2\) to 3 significant figures:

\[
V_2 \approx 6.28 \, \text{mL}
\]

So, the volume of the new container is:

\[
\boxed{6.28 \, \text{mL}}
\]