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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?

Physics

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Answer

B

Water changes from a liquid to a solid state

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Okay, Gezahagn, let's continue with these questions.

20. 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?

  1. 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)
    T2=0 °C=273.15 KT_2 = 0 \text{ °C} = 273.15 \text{ K}
    V2=?V_2 = ?

  2. Rearrange the formula to solve for V2V_2.

    V2=P1V1T2P2T1V_2 = \frac{P_1V_1T_2}{P_2T_1}
  3. Substitute the values and calculate V2V_2.

    V2=(3 atm)(25 L)(273.15 K)(1 atm)(290.15 K)V_2 = \frac{(3 \text{ atm})(25 \text{ L})(273.15 \text{ K})}{(1 \text{ atm})(290.15 \text{ K})} V2=20486.25 L290.15V_2 = \frac{20486.25 \text{ L}}{290.15} V2≈70.606 LV_2 \approx 70.606 \text{ L}
  4. Compare with the given options.

    The closest option is c. 70.60 L.
    The final answer is c. 70.60 L\boxed{\text{c. 70.60 L}}

    21. What is the molecular weight of a pure gaseous compound having a density of 4.95 g/L at -35 °C and 1020 torr?

  5. Identify the given values and the appropriate formula.

    We use the Ideal Gas Law in terms of density: PM=ρRTPM = \rho RT, where MM is the molecular weight.
    Given:
    Density (ρ\rho) = 4.95 g/L4.95 \text{ g/L}
    Temperature (TT) = −35 °C=−35+273.15=238.15 K-35 \text{ °C} = -35 + 273.15 = 238.15 \text{ K}
    Pressure (PP) = 1020 torr1020 \text{ torr}
    Gas constant (RR) = 62.36 L torr/(mol K)62.36 \text{ L torr/(mol K)} (since pressure is in torr)

  6. Rearrange the formula to solve for MM.

    M=ρRTPM = \frac{\rho RT}{P}
  7. Substitute the values and calculate MM.

    M=(4.95 g/L)(62.36 L torr/(mol K))(238.15 K)1020 torrM = \frac{(4.95 \text{ g/L})(62.36 \text{ L torr/(mol K)})(238.15 \text{ K})}{1020 \text{ torr}} M=73504.6981020 g/molM = \frac{73504.698}{1020} \text{ g/mol} M≈72.06 g/molM \approx 72.06 \text{ g/mol}
  8. Compare with the given options.

    The closest option is b. 72 g/mol.
    The final answer is b. 72 g/mol\boxed{\text{b. 72 g/mol}}

    22. If helium (He) diffuses through a porous barrier at a rate of 4.0 moles per minute, at what rate (in moles per minute) would oxygen (O₂) gas diffuse?

  9. Identify the given values and the appropriate law.

    We use Graham's Law of Diffusion: r1r2=M2M1\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}.
    Given:
    Rate of He (rHer_{He}) = 4.0 mol/min4.0 \text{ mol/min}
    Molar mass of He (MHeM_{He}) = 4.00 g/mol4.00 \text{ g/mol}
    Molar mass of O₂ (MO2M_{O_2}) = 2×16.00=32.00 g/mol2 \times 16.00 = 32.00 \text{ g/mol}
    We need to find the rate of O₂ (rO2r_{O_2}).

  10. Set up Graham's Law equation.

    rHerO2=MO2MHe\frac{r_{He}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{He}}}
  11. Substitute the values and solve for rO2r_{O_2}.

    4.0 mol/minrO2=32.00 g/mol4.00 g/mol\frac{4.0 \text{ mol/min}}{r_{O_2}} = \sqrt{\frac{32.00 \text{ g/mol}}{4.00 \text{ g/mol}}} 4.0rO2=8\frac{4.0}{r_{O_2}} = \sqrt{8} 4.0rO2≈2.828\frac{4.0}{r_{O_2}} \approx 2.828 rO2=4.02.828 mol/minr_{O_2} = \frac{4.0}{2.828} \text{ mol/min} rO2≈1.414 mol/minr_{O_2} \approx 1.414 \text{ mol/min}
  12. Compare with the given options.

    The closest option is a. 1.4.
    The final answer is a. 1.4\boxed{\text{a. 1.4}}

    23. What will happen water when freezes?
    The correct answer is b. Water changes from a liquid to a solid state.
    Freezing is the phase transition of a substance from a liquid to a solid state.

    24. What is sublimation?
    The correct answer is a. When a solid turns into a gas.
    Sublimation is the process where a substance changes directly from a solid to a gas without passing through the liquid phase.

    25. What eventually happens if energy is continually removed from a liquid?
    The correct answer is b. It freezes.
    Removing energy from a liquid causes its temperature to decrease. If enough energy is removed, the liquid will undergo a phase transition to a solid state, which is called freezing.

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