What is the pressure in millimeters of mercury of 0.0155 mole of He gas with a volume of 206 mL at a temperature of 30 degrees C?
I used the ideal gas law and ended up with 1.87 mmHg but it said that was wrong. Please help.
I am wondering
1) if you changed temp to kelvins
2 if you used the right gas constant (units in ml Kelvin mmhg)
Here's how I worked it:
P= ?
n= 0.0155 mol
v=206mL so .206L
t=30C +273 is 303K
then
0.0155mol *0.0821 L*atm/mol*K *303K divided by .206L which gives you 1.87 mmHg
What did I do wrong?
thanks! Duh! Got it! Yippie!
yep, you solved it in atm.
To solve this question, you correctly applied the ideal gas law, which is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
However, there seems to be an error in your calculation. Let's go through the solution step by step:
Step 1: Convert the temperature from degrees Celsius to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 30 + 273.15
T(K) = 303.15 K
Step 2: Convert the volume from mL to liters.
V(L) = V(mL) / 1000
V(L) = 206 mL / 1000
V(L) = 0.206 L
Step 3: Calculate the pressure using the ideal gas law.
P = (n x R x T) / V
Given:
n = 0.0155 moles (He gas)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 303.15 K (temperature)
V = 0.206 L (volume)
Plugging in the values:
P = (0.0155 mol x 0.0821 L·atm/(mol·K) x 303.15 K) / 0.206 L
P ≈ 1.88 atm
Now, to convert the pressure from atmospheres to millimeters of mercury (mmHg), we can use the fact that 1 atm = 760 mmHg.
Converting the pressure:
P(mmHg) = P(atm) x 760 mmHg / 1 atm
P(mmHg) ≈ 1.88 atm x 760 mmHg
P(mmHg) ≈ 1430 mmHg
Therefore, the correct value for the pressure in millimeters of mercury (mmHg) is approximately 1430 mmHg, not 1.87 mmHg as you calculated.
If you made a mistake during your calculations, please double-check your work to identify the error.