At what Kelvin temperature will 14.2 g of oxygen occupy a volume of 13.5 l at a pressure of 775 tour

Answer 1

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (775 torr)
V = volume (13.5 L)
n = number of moles
R = ideal gas constant (0.0821 L * atm / mol * K)
T = temperature in Kelvin

First, we need to find the number of moles of oxygen:

14.2 g O2 * (1 mol O2 / 32 g O2) = 0.44375 mol O2

Now we can plug in the values into the ideal gas law equation:

(775 torr) * (13.5 L) = (0.44375 mol) * (0.0821 L * atm / mol * K) * T

10462.5 = 0.036395 mol * T

T = 10462.5 / 0.036395
T = 287275 K

Therefore, the Kelvin temperature at which 14.2 g of oxygen will occupy a volume of 13.5 L at a pressure of 775 torr is approximately 287275 K.