An online conversion calculator for the units of pressure Pascals (Pa) and millimeters of mercury (mmHg) is presented. Examples and problems involving the conversion of Pa and mmHg are also included.
The Pascal and millimeter of mercury (mmHg) are units of pressure with the rate of conversion as follows:
\[ 1 \text{ mmHg} = 133.3224 \text{ Pa} \] \[ 1 \text{ Pa} = \left(\frac{1}{133.3224}\right) \text{ mmHg} \]Convert \( 2.1 \, \text{Pascals (Pa)} \) to mmHg and round the answer to 4 decimal places.
Given that \( 1 \, \text{Pa} = \left(\frac{1}{133.3224}\right) \text{mmHg} \):
\[ 2.1 \, \text{Pa} = 2.1 \times \left(\frac{1}{133.3224}\right) \text{mmHg} \approx 0.0157512916 \, \text{mmHg} \]Rounding to 4 decimal places gives:
\[ 2.1 \, \text{Pa} = 0.0158 \, \text{mmHg} \]Convert \( 694 \, \text{mmHg} \) to Pascals and round the answer to the nearest Pascal.
Given that \( 1 \, \text{mmHg} = 133.3224 \, \text{Pa} \):
\[ 694 \, \text{mmHg} = 694 \times 133.3224 \, \text{Pa} = 92,525.7456 \, \text{Pa} \]Rounding to the nearest Pascal (nearest unit) gives:
\[ 694 \, \text{mmHg} = 92,526 \, \text{Pa} \]Enter the number of Pascals (Pa) or millimeters of mercury (mmHg) to convert.
The atmospheric pressure (the pressure exerted by the weight of the atmosphere on everything on earth) is about \( 101,325 \, \text{Pa} \) at sea level. What is the atmospheric pressure in mmHg at sea level? Round the answer to the nearest unit.
Given that \( 1 \, \text{Pa} = \left(\frac{1}{133.3224}\right) \text{mmHg} \):
\[ 101,325 \, \text{Pa} = 101,325 \times \left(\frac{1}{133.3224}\right) \text{mmHg} \approx 759.99982 \, \text{mmHg} \]Rounding to the nearest unit (corrected calculation):
\[ 101,325 \, \text{Pa} = 760 \, \text{mmHg} \]The air pressure at ground level is about \( 724 \, \text{mmHg} \) and the air pressure at the cruising altitude of a commercial airplane is about \( 27,000 \, \text{Pa} \). What is the ratio \( R \) of the pressure on the ground to the pressure at cruising altitude? Round the answer to the nearest tenth. How many times is the pressure at ground level compared to the pressure at cruising altitude?
Given that \( 1 \, \text{mmHg} = 133.3224 \, \text{Pa} \), the air pressure at ground level in Pascals is:
\[ 724 \, \text{mmHg} = 724 \times 133.3224 \, \text{Pa} = 96,525.4176 \, \text{Pa} \]The ratio \( R \) is given by:
\[ R = \dfrac{96,525.4176 \, \text{Pa}}{27,000 \, \text{Pa}} \approx 3.575015 \]Rounding to the nearest tenth:
\[ R = 3.6 \]The pressure at ground level is 3.6 times the pressure at cruising altitude.
For a given temperature, the product of the pressure \( P \) and the volume \( V \) for a given mass of confined gas is constant (Boyle's law). Given an amount of a certain gas at a given temperature, at pressure \( P_1 = 250,000 \, \text{Pa} \) the volume of the gas is \( V_1 = 2 \, \text{liters} \). What pressure \( P_2 \) in mmHg is needed to compress the volume of the same gas under the same conditions to \( 1.3 \, \text{liters} \)? Round the answer to the nearest unit.
The product of pressure \( P_1 \) and volume \( V_1 \):
\[ P_1 \times V_1 = 250,000 \, \text{Pa} \times 2 \, \text{l} = 500,000 \, \text{Pa}\cdot\text{l} \]For a volume \( V_2 = 1.3 \, \text{l} \), the pressure \( P_2 \) satisfies:
\[ P_2 \times 1.3 = 500,000 \implies P_2 = \dfrac{500,000}{1.3} \approx 384,615.384615 \, \text{Pa} \]Convert \( P_2 \) to mmHg using \( 1 \, \text{Pa} = \left(\frac{1}{133.3224}\right) \text{mmHg} \):
\[ P_2 = 384,615.384615 \times \left(\frac{1}{133.3224}\right) \approx 2,884.851942 \, \text{mmHg} \]Rounding to the nearest unit:
\[ P_2 = 2,885 \, \text{mmHg} \]