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A uniform glass tube, closed at one end, contains a column of mercury 5 cm long. The length of the trapped air column is 26 cm when the tube is held vertically upright. When the tube is inverted, the

Physics

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b) Calculate the value of the atmospheric pressure?

  1. Identify the initial and final states of the air column.

    In Figure (a) (tube upright):
    Length of air column, L1=26 cmL_1 = 26 \text{ cm}
    Length of mercury thread, h=5 cmh = 5 \text{ cm}
    The pressure of the trapped air is the atmospheric pressure plus the pressure due to the mercury column:
    P1=Patm+h=Patm+5 cmHgP_1 = P_{atm} + h = P_{atm} + 5 \text{ cmHg}

    In Figure (b) (tube inverted):
    Length of air column, L2=30 cmL_2 = 30 \text{ cm}
    Length of mercury thread, h=5 cmh = 5 \text{ cm}
    The pressure of the trapped air is the atmospheric pressure minus the pressure due to the mercury column:
    P2=Patm−h=Patm−5 cmHgP_2 = P_{atm} - h = P_{atm} - 5 \text{ cmHg}

  2. Apply Boyle's Law.

    Assuming the temperature remains constant, Boyle's Law states P1V1=P2V2P_1V_1 = P_2V_2.
    Since the tube has a uniform cross-sectional area AA, the volume of the air column is V=LAV = LA.
    So, P1L1A=P2L2AP_1L_1A = P_2L_2A, which simplifies to P1L1=P2L2P_1L_1 = P_2L_2.

  3. Substitute the expressions for pressure and length into Boyle's Law.

    (Patm+5 cmHg)×26 cm=(Patm−5 cmHg)×30 cm(P_{atm} + 5 \text{ cmHg}) \times 26 \text{ cm} = (P_{atm} - 5 \text{ cmHg}) \times 30 \text{ cm}
  4. Solve for PatmP_{atm}.

    26Patm+26×5=30Patm−30×526 P_{atm} + 26 \times 5 = 30 P_{atm} - 30 \times 5 26Patm+130=30Patm−15026 P_{atm} + 130 = 30 P_{atm} - 150 130+150=30Patm−26Patm130 + 150 = 30 P_{atm} - 26 P_{atm} 280=4Patm280 = 4 P_{atm} Patm=2804P_{atm} = \frac{280}{4} Patm=70 cmHgP_{atm} = 70 \text{ cmHg}

    The value of the atmospheric pressure is 70 cmHg\boxed{\text{70 cmHg}}.

    c) A steel cylinder of capacity 0.5 m30.5 \text{ m}^3 contains nitrogen at a pressure of 30,000 Pa30,000 \text{ Pa} when the temperature is 27∘C27^\circ \text{C}. What will be the pressure of nitrogen if it is allowed to flow into another cylinder of capacity 9.5 m39.5 \text{ m}^3 with the temperature reduced to −23∘C-23^\circ \text{C}?

  5. Identify the initial conditions (P1,V1,T1P_1, V_1, T_1).

    Initial volume, V1=0.5 m3V_1 = 0.5 \text{ m}^3
    Initial pressure, P1=30,000 PaP_1 = 30,000 \text{ Pa}
    Initial temperature, T1=27∘CT_1 = 27^\circ \text{C}. Convert to Kelvin:
    T1=27+273.15=300.15 KT_1 = 27 + 273.15 = 300.15 \text{ K}

  6. Identify the final conditions (V2,T2V_2, T_2) and the unknown (P2P_2).

    The gas flows into another cylinder of capacity 9.5 m39.5 \text{ m}^3. The total final volume is the sum of the capacities of both cylinders:
    Final volume, V2=0.5 m3+9.5 m3=10 m3V_2 = 0.5 \text{ m}^3 + 9.5 \text{ m}^3 = 10 \text{ m}^3
    Final temperature, T2=−23∘CT_2 = -23^\circ \text{C}. Convert to Kelvin:
    T2=−23+273.15=250.15 KT_2 = -23 + 273.15 = 250.15 \text{ K}
    Final pressure, P2=?P_2 = ?

  7. Apply the Combined Gas Law.

    The Combined Gas Law is given by:

    P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}
  8. Substitute the values and solve for P2P_2.

    (30,000 Pa)(0.5 m3)300.15 K=P2(10 m3)250.15 K\frac{(30,000 \text{ Pa})(0.5 \text{ m}^3)}{300.15 \text{ K}} = \frac{P_2(10 \text{ m}^3)}{250.15 \text{ K}} P2=(30,000 Pa)(0.5 m3)(250.15 K)(300.15 K)(10 m3)P_2 = \frac{(30,000 \text{ Pa})(0.5 \text{ m}^3)(250.15 \text{ K})}{(300.15 \text{ K})(10 \text{ m}^3)} P2=15000×250.15300.15×10P_2 = \frac{15000 \times 250.15}{300.15 \times 10} P2=37522503001.5P_2 = \frac{3752250}{3001.5} P2≈1250.12 PaP_2 \approx 1250.12 \text{ Pa}

    The pressure of nitrogen will be 1250.12 Pa\boxed{\text{1250.12 Pa}}.

    1. a) State Archimedes principle. Archimedes' principle states that when an object is wholly or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by the object.

    b) State one way of making a hydrometer more sensitive.
    One way to make a hydrometer more sensitive is to increase the length of its stem or decrease the diameter of its stem.

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