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To convert the cabin pressure from millimeters of mercury (mmHg) to atmospheres (atm), follow these steps:
1. Identify the Given Pressure: The cabin pressure is provided as [tex]\(6.00 \times 10^2 \, \text{mmHg}\)[/tex].
2. Understand the Conversion Factor: We know that [tex]\(1 \, \text{atm} = 760 \, \text{mmHg}\)[/tex]. Therefore, to convert from mmHg to atm, we use the conversion factor where [tex]\(1 \, \text{mmHg} = \frac{1}{760} \, \text{atm}\)[/tex].
3. Set Up the Conversion Calculation:
[tex]\[ P(\text{atm}) = P(\text{mmHg}) \times \left( \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \right) \][/tex]
4. Apply the Given Values:
[tex]\[ P(\text{atm}) = 6.00 \times 10^2 \, \text{mmHg} \times \left( \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \right) \][/tex]
5. Calculate the Pressure in Atmospheres:
[tex]\[ P(\text{atm}) = 600 \, \text{mmHg} \times \frac{1}{760} \, \text{atm} \][/tex]
By performing the division:
[tex]\[ P(\text{atm}) \approx 0.7894736842105263 \, \text{atm} \][/tex]
So, the pressure inside the cabin in atmospheres is approximately [tex]\(0.789 \, \text{atm}\)[/tex] (Typically rounded to three significant figures, this would be [tex]\(0.789 \, \text{atm}\)[/tex]).
Therefore, the pressure inside the cabin is:
[tex]\[ P = 0.789 \, \text{atm} \][/tex]
1. Identify the Given Pressure: The cabin pressure is provided as [tex]\(6.00 \times 10^2 \, \text{mmHg}\)[/tex].
2. Understand the Conversion Factor: We know that [tex]\(1 \, \text{atm} = 760 \, \text{mmHg}\)[/tex]. Therefore, to convert from mmHg to atm, we use the conversion factor where [tex]\(1 \, \text{mmHg} = \frac{1}{760} \, \text{atm}\)[/tex].
3. Set Up the Conversion Calculation:
[tex]\[ P(\text{atm}) = P(\text{mmHg}) \times \left( \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \right) \][/tex]
4. Apply the Given Values:
[tex]\[ P(\text{atm}) = 6.00 \times 10^2 \, \text{mmHg} \times \left( \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \right) \][/tex]
5. Calculate the Pressure in Atmospheres:
[tex]\[ P(\text{atm}) = 600 \, \text{mmHg} \times \frac{1}{760} \, \text{atm} \][/tex]
By performing the division:
[tex]\[ P(\text{atm}) \approx 0.7894736842105263 \, \text{atm} \][/tex]
So, the pressure inside the cabin in atmospheres is approximately [tex]\(0.789 \, \text{atm}\)[/tex] (Typically rounded to three significant figures, this would be [tex]\(0.789 \, \text{atm}\)[/tex]).
Therefore, the pressure inside the cabin is:
[tex]\[ P = 0.789 \, \text{atm} \][/tex]
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