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13 August, 06:17

A 2.50 L flask was used to collect a 1.65 g sample of propane gas,. After the sample was collected, the gas pressure was found to be 736 mmHg. What was the temperature of the propane in the flask

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Answers (2)
  1. 13 August, 06:36
    0
    788.6K

    Explanation:

    First, let us calculate the number of mole propane (C3H8). This can be achieved as shown below:

    Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

    Mass of C3H8 obtained from the question = 1.65g

    Number of mole = Mass / Molar Mass

    Number of mole of C3H8 = 1.65/44 = 0.0375mol

    Now we can obtain the temperature of propane gas as shown below:

    V = 2.50L

    P = 736mmHg

    Recall that 760mmHg = 1atm

    Therefore 736mmHg = 736/760 = 0.97atm

    n = 0.0375mol

    R = 0.082atm. L/Kmol

    T = ?

    PV = nRT

    T = PV/nR

    T = 0.97x 2.5 / 0.0375x0.082

    T = 788.6K

    The temperature of the propane was 788.6K
  2. 13 August, 06:37
    0
    The temperature of propane in the flask was 788.9 K or 515.75 °C

    Explanation:

    Step 1: Data given

    Volume of the flask = 2.50 L

    Mass of propane = 1.65 grams

    Molar mass of propane = 44.1 g/mol

    Pressure = 736 mmHg = 0.968421 atm

    Step 2: Calculate moles propane

    Moles propane = mass propane / molar mass propane

    Moles propane = 1.65 grams / 44.1 g/mol

    Moles propane = 0.0374 moles

    Step 3: Calculate temperature

    p*V = n*R*T

    T = (p*V) / (n*R)

    ⇒with T = the temperature in the flask

    ⇒with p = the pressure of propane gas = 0.968421 atm

    ⇒with V = the volume of the flask = 2.50 L

    ⇒with n = the number of moles propane gas = 0.0374 moles

    ⇒with R = the gas constant = 0.08206 L*atm/mol*K

    T = (0.968421 * 2.50) / (0.0374*0.08206)

    T = 788.9 K

    The temperature of propane in the flask was 788.9 K or 515.75 °C
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