To find the volume of the gas-filled balloon at Standard Temperature and Pressure (STP), we'll use the Combined Gas Law. The Combined Gas Law is given by:
[tex]\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\][/tex]
where:
- [tex]\(P_1\)[/tex], [tex]\(V_1\)[/tex], and [tex]\(T_1\)[/tex] are the initial pressure, volume, and temperature, respectively.
- [tex]\(P_2\)[/tex], [tex]\(V_2\)[/tex], and [tex]\(T_2\)[/tex] are the pressure, volume, and temperature at the final conditions (STP).
First, let's list the given values:
- Initial volume, [tex]\( V_1 = 5.8 \text{ L} \)[/tex]
- Initial temperature, [tex]\( T_1 = 20.7^\circ\text{C} \)[/tex]
- Initial pressure, [tex]\( P_1 = 140.7 \text{ kPa} \)[/tex]
At STP, the standard conditions are:
- Standard Temperature, [tex]\( T_2 = 0^\circ\text{C} = 273.15 \text{ K} \)[/tex]
- Standard Pressure, [tex]\( P_2 = 101.325 \text{ kPa} \)[/tex]
First, convert the initial temperature from Celsius to Kelvin:
[tex]\[
T_1 = 20.7 + 273.15 = 293.85 \text{ K}
\][/tex]
Now, let's apply the Combined Gas Law to solve for [tex]\(V_2\)[/tex]:
[tex]\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\][/tex]
Rearranging to solve for [tex]\(V_2\)[/tex]:
[tex]\[
V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}
\][/tex]
Plugging in the values:
[tex]\[
V_2 = \frac{140.7 \text{ kPa} \times 5.8 \text{ L} \times 273.15 \text{ K}}{101.325 \text{ kPa} \times 293.85 \text{ K}}
\][/tex]
Performing the calculations:
[tex]\[
V_2 = \frac{140.7 \times 5.8 \times 273.15}{101.325 \times 293.85}
\][/tex]
[tex]\[
V_2 \approx 7.49 \text{ L}
\][/tex]
So, the volume of the balloon at STP will be approximately 7.49 liters.
