A mass of oxygen occurps 7.0 millimetres under the pressure of 740 torr deretimen the same volume of Mass of gas and standard pressure and the temperature remaining constant

Physics

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Answer

6.8 mL\text{6.8 mL}

Step-by-step solution

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This problem involves a gas undergoing a change in pressure while its temperature remains constant. This scenario is described by Boyle's Law, which states that for a fixed mass of gas at constant temperature, the pressure (PP) and volume (VV) are inversely proportional. Mathematically, this is expressed as:

P1V1=P2V2P_1 V_1 = P_2 V_2

Where:

  • P1P_1 is the initial pressure
  • V1V_1 is the initial volume
  • P2P_2 is the final pressure
  • V2V_2 is the final volume

  1. Identify the given values and make necessary assumptions.

    • Initial pressure, P1=740 torrP_1 = 740 \text{ torr}
    • Initial volume, V1=7.0 mLV_1 = 7.0 \text{ mL} (Assuming "7.0 millimetres" is a typo for "7.0 milliliters", as millimeters is a unit of length, not volume, in this context.)
    • Final pressure, P2=760 torrP_2 = 760 \text{ torr} (Standard pressure is commonly taken as 1 atmosphere, which is equal to 760 torr.)
    • Temperature remains constant.
    • We need to find the final volume, V2V_2.

  2. Rearrange Boyle's Law to solve for V2V_2.

    V2=P1V1P2V_2 = \frac{P_1 V_1}{P_2}

  3. Substitute the known values into the equation.

    V2=(740 torr)×(7.0 mL)760 torrV_2 = \frac{(740 \text{ torr}) \times (7.0 \text{ mL})}{760 \text{ torr}}

  4. Calculate the final volume.

    V2=5180 torr⋅mL760 torrV_2 = \frac{5180 \text{ torr} \cdot \text{mL}}{760 \text{ torr}} V2≈6.815789 mLV_2 \approx 6.815789 \text{ mL}

  5. Round the answer to an appropriate number of significant figures.

    Since the initial volume (7.0 mL) has two significant figures, we round our answer to two significant figures.

    V2≈6.8 mLV_2 \approx 6.8 \text{ mL}

    The volume of the oxygen gas at standard pressure (760 torr) and constant temperature is approximately 6.8 mL\boxed{\text{6.8 mL}}.

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