An open-ended mercury manometer is used to measure the pressure exerted by a trapped gas as shown in the figure. Atmospheric pressure is 750. mmHg. What is the pressure (in mmHg or torr) of the trapped gas if h =23 cm?
I got 893.462, but that wasn't right. Can someone help please?
To determine the pressure of the trapped gas using the open-ended mercury manometer, you need to consider the difference in heights between the two sides of the manometer.
Given that the height of the mercury column on one side (h) is 23 cm, we can use this information to calculate the pressure of the trapped gas.
First, we need to convert the height of the mercury column from cm to mm, as the unit of pressure used in this question is mmHg.
1 cm = 10 mm
Therefore, the height of the mercury column in mm = 23 cm x 10 mm/cm = 230 mm.
Next, we need to find the difference in height between the two sides of the mercury manometer. Since one side of the manometer is open to the atmosphere, the height of the column on the other side will represent the pressure of the trapped gas.
The pressure exerted by the trapped gas can be calculated using the equation:
Pressure = Atmospheric pressure + (Height of the mercury column on the closed side - Height of the mercury column on the open side)
Given that the atmospheric pressure is 750 mmHg and the height on the open side is 0 mmHg, the equation becomes:
Pressure = 750 mmHg + (Height of the mercury column on the closed side - 0 mmHg)
Substituting the given value of the height on the closed side (230 mm) into the equation:
Pressure = 750 mmHg + (230 mm - 0 mmHg)
Pressure = 750 mmHg + 230 mmHg
Pressure = 980 mmHg
Thus, the pressure of the trapped gas is 980 mmHg or 980 torr.