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Question 1: State Avogadro's Law and solve this problem. A gas occupies 12 dm³ at constant temperature and pressure. Calculate the volume occupied by 5 moles if 2 moles occupy 12 dm³.

Physics

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2 parts
1

30 dm3\text{30 dm}^3

2

300 kPa\text{300 kPa}

Step-by-step solution

8 steps

Here are the solutions to two questions from Section B:

Part 1· 4 steps

State Avogadro's Law and solve this problem: A gas occupies 12 dm312 \text{ dm}^3 at constant temperature and pressure. Calculate the volume occupied by 5 moles if 2 moles occupy 12 dm312 \text{ dm}^3.

  1. State Avogadro's Law.

    Avogadro's Law states that, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas. Mathematically, this can be expressed as V∝nV \propto n or V1n1=V2n2\frac{V_1}{n_1} = \frac{V_2}{n_2}.

  2. Identify the given values for the problem.

    Initial volume (V1V_1) = 12 dm312 \text{ dm}^3
    Initial number of moles (n1n_1) = 2 moles2 \text{ moles}
    Final number of moles (n2n_2) = 5 moles5 \text{ moles}
    Final volume (V2V_2) = ?

  3. Apply Avogadro's Law formula and substitute the values.

    V1n1=V2n2\frac{V_1}{n_1} = \frac{V_2}{n_2} 12 dm32 moles=V25 moles\frac{12 \text{ dm}^3}{2 \text{ moles}} = \frac{V_2}{5 \text{ moles}}
  4. Solve for V2V_2.

    V2=12 dm3×5 moles2 molesV_2 = \frac{12 \text{ dm}^3 \times 5 \text{ moles}}{2 \text{ moles}} V2=60 dm32V_2 = \frac{60 \text{ dm}^3}{2} V2=30 dm3V_2 = 30 \text{ dm}^3

    The volume occupied by 5 moles of the gas is 30 dm3\boxed{\text{30 dm}^3}.

Part 2

300 kPa\text{300 kPa}

State Gay-Lussac's Law and calculate the final pressure of a gas initially at 150 kPa150 \text{ kPa} and 300 K300 \text{ K} when heated to 600 K600 \text{ K} at constant volume.

  1. State Gay-Lussac's Law.

    Gay-Lussac's Law (also known as Amontons's Law) states that for a fixed mass and volume of gas, the pressure is directly proportional to its absolute (Kelvin) temperature. Mathematically, this can be expressed as P∝TP \propto T or P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}.

  2. Identify the given values for the problem.

    Initial pressure (P1P_1) = 150 kPa150 \text{ kPa}
    Initial temperature (T1T_1) = 300 K300 \text{ K}
    Final temperature (T2T_2) = 600 K600 \text{ K}
    Final pressure (P2P_2) = ?

  3. Apply Gay-Lussac's Law formula and substitute the values.

    P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2} 150 kPa300 K=P2600 K\frac{150 \text{ kPa}}{300 \text{ K}} = \frac{P_2}{600 \text{ K}}
  4. Solve for P2P_2.

    P2=150 kPa×600 K300 KP_2 = \frac{150 \text{ kPa} \times 600 \text{ K}}{300 \text{ K}} P2=90000 kPa300P_2 = \frac{90000 \text{ kPa}}{300} P2=300 kPaP_2 = 300 \text{ kPa}

    The final pressure of the gas is 300 kPa\boxed{\text{300 kPa}}.

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