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Describe the properties of solids, liquids, and gases. Include definitions for states of matter and related concepts like sublimation, absolute zero, and Avogadro's Law.

Chemistry

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Answer

A

Gas

Step-by-step solution

10 steps

Part i

Write what each of these refers to.

  1. They have a definite shape and a definite volume. Solid
  2. Their molecules are highly disordered. Gas
  3. The motion of their molecules is highly restricted. Solid
  4. They can be easily compressed. Gas
  5. They have a tendency to flow. Liquid, Gas
  6. They can take the shape of their container. Liquid, Gas
  7. They can move in all direction at high speed. Gas
  8. They can easily diffuse through each other. Gas
  9. They can sublime. Solid
  10. They have less density relative to the other states. Gas

Part ii

Complete the following statements.

  1. Liquids and gases have no definite shape.
  2. The freezing point of a liquid is the same as the melting point of its solid.
  3. The temperature at which a crystalline solid is directly converted to a gas is called the sublimation point.
  4. The lowest attainable temperature is absolute zero.
  5. "At the fixed temperature and pressure, equal volumes of different gases have equal numbers of molecules." This statement is postulated by Avogadro's Law.

Part iii· 10 steps

Multiple Choice Questions.

  1. Which of the following would take up the entire shape of the container it is enclosed within? The correct answer is a. Gas. Gases expand to fill the entire volume and shape of their container.

  2. If a gas expands at constant temperature, it indicates that: The correct answer is d. Kinetic energy of molecules remains the same. Temperature is a measure of the average kinetic energy of gas molecules. If the temperature is constant, the average kinetic energy of the molecules remains the same.

  3. A sample of oxygen occupies 47.2 L under a pressure of 1240 torr at 25 °C. What volume would it occupy at 25 °C if the pressure were decreased to 730 torr?

  1. Identify the given values and the law to use.

    Since the temperature is constant, we use Boyle's Law: P1V1=P2V2P_1V_1 = P_2V_2.
    Given:
    P1=1240 torrP_1 = 1240 \text{ torr}
    V1=47.2 LV_1 = 47.2 \text{ L}
    P2=730 torrP_2 = 730 \text{ torr}
    V2=?V_2 = ?

  2. Rearrange the formula to solve for V2V_2.

    V2=P1V1P2V_2 = \frac{P_1V_1}{P_2}
  3. Substitute the values and calculate V2V_2.

    V2=1240 torr×47.2 L730 torrV_2 = \frac{1240 \text{ torr} \times 47.2 \text{ L}}{730 \text{ torr}} V2=58528 L730V_2 = \frac{58528 \text{ L}}{730} V2≈80.175 LV_2 \approx 80.175 \text{ L}
  4. Compare with the given options.

    The closest option is d. 80.2 L.
    The final answer is d. 80.2 L\boxed{\text{d. 80.2 L}}

    1. A sample of nitrogen occupies 5.50 liters under a pressure of 900.0 torr at 25.0 °C. At what temperature will it occupy 10.0 liters at the same pressure?
  5. Identify the given values and the law to use.

    Since the pressure is constant, we use Charles's Law: V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}.
    Temperatures must be in Kelvin.
    Given:
    V1=5.50 LV_1 = 5.50 \text{ L}
    T1=25.0 °C=25.0+273.15=298.15 KT_1 = 25.0 \text{ °C} = 25.0 + 273.15 = 298.15 \text{ K}
    V2=10.0 LV_2 = 10.0 \text{ L}
    T2=?T_2 = ?

  6. Rearrange the formula to solve for T2T_2.

    T2=V2T1V1T_2 = \frac{V_2T_1}{V_1}
  7. Substitute the values and calculate T2T_2.

    T2=10.0 L×298.15 K5.50 LT_2 = \frac{10.0 \text{ L} \times 298.15 \text{ K}}{5.50 \text{ L}} T2=2981.5 K5.50T_2 = \frac{2981.5 \text{ K}}{5.50} T2≈542.09 KT_2 \approx 542.09 \text{ K}
  8. Convert T2T_2 back to Celsius.

    T2 (°C)=542.09−273.15=268.94 °CT_2 \text{ (°C)} = 542.09 - 273.15 = 268.94 \text{ °C}
  9. Compare with the given options.

    The closest option is b. 269 °C.
    The final answer is b. 269 °C\boxed{\text{b. 269 °C}}

    1. A quantity of gas has a volume of 25 L at 17 °C and 3 atm of pressure. To what volume must the gas be increased for a gas to be under STP conditions?
  10. Identify the given values and the law to use.

    This is a Combined Gas Law problem: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}.
    Temperatures must be in Kelvin. STP conditions are P2=1 atmP_2 = 1 \text{ atm} and T2=0 °C=273.15 KT_2 = 0 \text{ °C} = 273.15 \text{ K}.
    Given:
    P1=3 atmP_1 = 3 \text{ atm}
    V1=25 LV_1 = 25 \text{ L}
    T1=17 °C=17+273.15=290.15 KT_1 = 17 \text{ °C} = 17 + 273.15 = 290.15 \text{ K}
    P2=1 atmP_2 = 1 \text{ atm} (STP)
    $T_2 = 0 °C = 273.15 \text{

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