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A gas has a volume of 146 cm3 at 18°C and 0.971 atm. What is its volume at STP?

Physics

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Answer

133 cm3\text{133 cm}^3

Step-by-step solution

5 steps

To find the volume of the gas at Standard Temperature and Pressure (STP), we will use the Combined Gas Law. STP is defined as 0∘C0^\circ\text{C} (273 K) and 1 atm1 \text{ atm}.

  1. List the initial conditions (P1,V1,T1P_1, V_1, T_1).

    • Initial volume (V1V_1) = 146 cm3146 \text{ cm}^3
    • Initial temperature (T1T_1) = 18∘C18^\circ\text{C}
    • Initial pressure (P1P_1) = 0.971 atm0.971 \text{ atm}
  2. Convert the initial temperature to Kelvin.

    T1=18∘C+273.15=291.15 KT_1 = 18^\circ\text{C} + 273.15 = 291.15 \text{ K}
  3. List the final conditions at STP (P2,T2P_2, T_2).

    • Standard temperature (T2T_2) = 0∘C=273.15 K0^\circ\text{C} = 273.15 \text{ K}
    • Standard pressure (P2P_2) = 1 atm1 \text{ atm}
    • Final volume (V2V_2) = ?
  4. Apply the Combined Gas Law.

    The Combined Gas Law states:

    P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

    Rearrange the formula to solve for V2V_2:

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

    V2=(0.971 atm)×(146 cm3)×(273.15 K)(1 atm)×(291.15 K)V_2 = \frac{(0.971 \text{ atm}) \times (146 \text{ cm}^3) \times (273.15 \text{ K})}{(1 \text{ atm}) \times (291.15 \text{ K})} V2=38758.9999 cm3291.15V_2 = \frac{38758.9999 \text{ cm}^3}{291.15} V2≈133.12 cm3V_2 \approx 133.12 \text{ cm}^3

    Rounding to three significant figures:

    V2≈133 cm3V_2 \approx 133 \text{ cm}^3

    The volume of the gas at STP is:

    133 cm3\text{133 cm}^3
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