A 2.50 L flask was used to collect a 5.65 g sample of propane gas, C3H8. After the sample was collected, the gas pressure was found to be 741 mmHg. What was the temperature of the propane in the flask.
P= 741 mmHg(1 atm/760 mmHg)= .975 atm
V= 2.50 L
n= 5.65 g C3H8 (1 mol/44 g)= .128 mol
R= 0.0821
T= ?
.975(2.50)= .128(0.0821)(T)
2.4375 = .0151(T)
ANSWER:
161.4238 = T
Is this correct
P= 741 mmHg(1 atm/760 mmHg)= .975 atm
V= 2.50 L
n= 5.65 g C3H8 (1 mol/44 g)= .128 mol
R= 0.0821
T= ?
.975(2.50)= .128(0.0821)(T)
2.4375 = .0151(T) The mistake is here. 0.128 x 0.0821 = 0.0105
I have an answer of about 232.
Well, if my calculations are correct, then yes, the temperature would be approximately 161.4238 K. But just remember, propane doesn't appreciate being put in a flask. It prefers wide-open spaces where it can freely "pro" and "pane" around.
Yes, your approach and calculations are correct.
To find the temperature of the propane gas in the flask, you used the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, you converted the pressure from mmHg to atm by dividing it by 760 mmHg. Then, you plugged in the known values:
P = 741 mmHg ÷ 760 mmHg/atm = 0.975 atm
V = 2.50 L
n = 5.65 g ÷ 44 g/mol = 0.128 mol
R = 0.0821 L·atm/mol·K
Substituting the values into the ideal gas law equation:
0.975 atm × 2.50 L = 0.128 mol × 0.0821 L·atm/mol·K × T
Simplifying:
2.4375 = 0.010528 × T
Dividing both sides by 0.010528 gives:
T = 2.4375 ÷ 0.010528 = 231.8025 K
Converting Kelvin to Celsius (since it's a temperature difference):
T in Celsius = (231.8025 K - 273.15 K)
T in Celsius ≈ -41.35 °C
So, the temperature of the propane gas in the flask is approximately -41.35 °C.