Connect with knowledgeable individuals and get your questions answered on IDNLearn.com. Our platform offers reliable and detailed answers, ensuring you have the information you need.
Sagot :
Certainly! Let's solve the problem step-by-step using the Combined Gas Law which relates pressure (P), volume (V), and temperature (T) of a gas sample with the equation:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Here are the given values:
- Initial volume [tex]\( V_1 = 54.0 \)[/tex] L
- Initial pressure [tex]\( P_1 = 759 \)[/tex] mmHg
- Initial temperature [tex]\( T_1 = 21.1^{\circ} \text{C} \)[/tex]
First, we need to convert the initial temperature from Celsius to Kelvin:
[tex]\[ T_1 = 21.1 + 273.15 = 294.25 \text{ K} \][/tex]
Next, we are given the conditions at the higher altitude:
- Temperature [tex]\( T_2 = -3.86^{\circ} \text{C} \)[/tex]
- Pressure [tex]\( P_2 = 0.0757 \text{ atm} \)[/tex]
Again, convert the temperature [tex]\( T_2 \)[/tex] to Kelvin:
[tex]\[ T_2 = -3.86 + 273.15 = 269.29 \text{ K} \][/tex]
Also, convert the pressure [tex]\( P_2 \)[/tex] into mmHg (since 1 atm = 760 mmHg):
[tex]\[ P_2 = 0.0757 \times 760 = 57.532 \text{ mmHg} \][/tex]
Now, we are ready to solve for the new volume [tex]\( V_2 \)[/tex] using the Combined Gas Law:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Rearranging the formula to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} \][/tex]
Substituting the given values into the equation:
[tex]\[ V_2 = \frac{759 \,\text{mmHg} \times 54.0 \,\text{L} \times 269.29 \,\text{K}}{294.25 \,\text{K} \times 57.532 \,\text{mmHg}} \][/tex]
[tex]\[ V_2 = 651.97 \, \text{L} \quad \text{(rounded to two decimal places)} \][/tex]
Therefore, the volume of the weather balloon at the higher altitude is:
[tex]\[ V = 651.97 \, \text{L} \][/tex]
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Here are the given values:
- Initial volume [tex]\( V_1 = 54.0 \)[/tex] L
- Initial pressure [tex]\( P_1 = 759 \)[/tex] mmHg
- Initial temperature [tex]\( T_1 = 21.1^{\circ} \text{C} \)[/tex]
First, we need to convert the initial temperature from Celsius to Kelvin:
[tex]\[ T_1 = 21.1 + 273.15 = 294.25 \text{ K} \][/tex]
Next, we are given the conditions at the higher altitude:
- Temperature [tex]\( T_2 = -3.86^{\circ} \text{C} \)[/tex]
- Pressure [tex]\( P_2 = 0.0757 \text{ atm} \)[/tex]
Again, convert the temperature [tex]\( T_2 \)[/tex] to Kelvin:
[tex]\[ T_2 = -3.86 + 273.15 = 269.29 \text{ K} \][/tex]
Also, convert the pressure [tex]\( P_2 \)[/tex] into mmHg (since 1 atm = 760 mmHg):
[tex]\[ P_2 = 0.0757 \times 760 = 57.532 \text{ mmHg} \][/tex]
Now, we are ready to solve for the new volume [tex]\( V_2 \)[/tex] using the Combined Gas Law:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Rearranging the formula to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} \][/tex]
Substituting the given values into the equation:
[tex]\[ V_2 = \frac{759 \,\text{mmHg} \times 54.0 \,\text{L} \times 269.29 \,\text{K}}{294.25 \,\text{K} \times 57.532 \,\text{mmHg}} \][/tex]
[tex]\[ V_2 = 651.97 \, \text{L} \quad \text{(rounded to two decimal places)} \][/tex]
Therefore, the volume of the weather balloon at the higher altitude is:
[tex]\[ V = 651.97 \, \text{L} \][/tex]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.