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

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

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

is there multiple choice answers?

yes

explanation:

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