Question

1. given a barometric pressure of 762.4 mmhg, calculate the pressure of each gas sample as indicated by the manometer. missed this? read section 6.2 (a) (b) 32. given a barometric pressure of 751...

Explanation:

Step1: Analyze Manometer (a)

In manometer (a), the gas pressure \( P_{\text{gas}} \) and atmospheric pressure \( P_{\text{atm}} \) relate as \( P_{\text{gas}} = P_{\text{atm}} - h \). First, convert \( h \) from cm to mm: \( h = 4.0 \, \text{cm} = 40.0 \, \text{mm} \) (since 1 cm = 10 mm). Given \( P_{\text{atm}} = 762.4 \, \text{mmHg} \), so \( P_{\text{gas}} = 762.4 - 40.0 = 722.4 \, \text{mmHg} \).

Step2: Analyze Manometer (b)

In manometer (b), the gas pressure \( P_{\text{gas}} = P_{\text{atm}} + h \). Convert \( h = 4.0 \, \text{cm} = 40.0 \, \text{mm} \). Then \( P_{\text{gas}} = 762.4 + 40.0 = 802.4 \, \text{mmHg} \). (Wait, wait—wait, the scale: looking at the manometer (a) and (b), maybe the \( h \) is 4.0 cm? Wait, the diagram: in (a), the mercury level is lower on the gas side? Wait, no—wait, the first manometer (a): the atmospheric pressure is pushing up, so gas pressure is less than atm. The second (b): gas is pushing down, so gas pressure is more than atm. Wait, maybe the \( h \) is 4.0 cm? Wait, the scale: each cm is marked, so from 0 to 4 cm, so \( h = 4.0 \, \text{cm} = 40 \, \text{mm} \). Wait, but maybe the reading is 4.0 cm? Wait, let's recheck.

Wait, the problem says "barometric pressure of 762.4 mmHg". For manometer (a): the gas is in the bulb, connected to a U-tube. The atmospheric pressure is on one side, gas on the other. If the mercury level is 4.0 cm (40 mm) higher on the atmospheric side, then gas pressure \( P_{\text{gas}} = P_{\text{atm}} - h \). So \( h = 4.0 \, \text{cm} = 40 \, \text{mm} \). So \( 762.4 - 40 = 722.4 \, \text{mmHg} \). For manometer (b): the mercury level is 4.0 cm higher on the gas side, so \( P_{\text{gas}} = P_{\text{atm}} + h = 762.4 + 40 = 802.4 \, \text{mmHg} \). Wait, but maybe the \( h \) is 4.0 cm? Wait, the scale: the manometer has markings from -4 to +4 cm? Wait, no, the diagram: (a) has the atmospheric pressure side at 4 cm? Wait, maybe I misread. Wait, the first manometer (a): the arrow is at 4 cm? Wait, the vertical scale is in cm, with 0 at the middle. So the difference \( h \) is 4.0 cm (from 0 to 4 cm, so 40 mm). So that's correct.

Wait, but maybe the question is to find the pressure for each? Let's confirm the formula for U-tube manometers:

• If the gas side mercury is lower: \( P_{\text{gas}} = P_{\text{atm}} + h \)
• If the gas side mercury is higher: \( P_{\text{gas}} = P_{\text{atm}} - h \)

In (a): gas side mercury is higher (since atmospheric pressure is pushing up, so gas pressure is less, so mercury is higher on gas side). So \( h = 4.0 \, \text{cm} = 40 \, \text{mm} \). So \( P_{\text{gas}} = 762.4 - 40 = 722.4 \, \text{mmHg} \).

In (b): gas side mercury is lower (gas is pushing down, so mercury is lower on gas side), so \( P_{\text{gas}} = 762.4 + 40 = 802.4 \, \text{mmHg} \).

Wait, but maybe the \( h \) is 4.0 cm? Let's check the units. 1 cm = 10 mm, so 4 cm = 40 mm. So that's correct.

So for (a): \( 762.4 - 40 = 722.4 \, \text{mmHg} \)

For (b): \( 762.4 + 40 = 802.4 \, \text{mmHg} \)

Wait, but maybe the \( h \) is 4.0 cm? Let's confirm the diagram. The first manometer (a) has the mercury level at 4 cm on the atmospheric side, so the difference is 4 cm. So yes, 40 mm.

Answer:

For manometer (a): \(\boldsymbol{722.4 \, \text{mmHg}}\)
For manometer (b): \(\boldsymbol{802.4 \, \text{mmHg}}\)