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