0.000014 mol of argon gas is placed in a 600 mL light bulb. What's the pressure of the gas in the bulb at 23C?
PV=nRT
(0.6L) X = (0.000014 mol) (0.08206 L-atm/mol-K)(296.15 K)
X= 5.7 *10^-4 atm ?
Depending upon the 600 number (is that 600. or 6 x 10^2)-----
If 600. then the answer is 5.67 x 10^-4 atm. If 6 x 10^2, then the answer is 6 x 10^-4.
Yes
I don't understand... 600 is the same as 6*10^2.
"600 mL" was the original volume given in the question.
No, 6 x 10^2 has one significant figure, the 6, while 600. has 3 s.f. (Technically, 600 has only one since we don't know that the two zeros are significant. Placing the decimal after the second zero lets us know that 600 has three s.f.) To avoid confusion, we write 600. to mean three s.f. and we write 6 x 10^2 to mean 1 s.f.
To find the pressure of the gas in the bulb, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
First, let's convert the volume of the light bulb to liters:
600 mL = 600/1000 L = 0.6 L
Next, convert the temperature from Celsius to Kelvin:
T = 23°C + 273.15 = 296.15 K
Now, we can substitute the values into the equation and solve for the pressure (P):
P * 0.6 L = (0.000014 mol) * (0.08206 L-atm/mol-K) * (296.15 K)
P * 0.6 L = 0.010994116
Divide both sides of the equation by 0.6 L:
P = 0.010994116 / 0.6
P = 0.0183 atm
So the pressure of the gas in the bulb at 23°C is approximately 0.0183 atm.