(a) An ideal gas occupies a volume of 1.0 cm3 at 20°C and atmospheric pressure. Determine the number of molecules of gas in the container.
i got that right to be 2.5e19 molecules
(b) If the pressure of the 1.0 cm3 volume is reduced to 1.0 10-11 Pa (an extremely good vacuum) while the temperature remains constant, how many moles of gas remain in the container?
im not sure how to do this part. and i keep getting it wrong. can you please help me?
Sure, I'd be happy to help you with this question! But first, let me lighten the mood with a joke:
Why did the scarecrow win an award?
Because he was outstanding in his field!
Now, let's tackle the question. To determine the number of moles of gas remaining in the container, we need to use the ideal gas law equation: PV = nRT.
Since the volume (V), temperature (T), and gas constant (R) remain constant, we can simplify the equation to P = nR/T.
Given that the pressure (P) is 1.0 × 10-11 Pa, we can substitute that value into the equation and solve for n (the number of moles):
n = P × V / (R × T)
But before we do that, let's convert the volume from cubic centimeters (cm3) to liters (L) to match the units used in the ideal gas law equation. Since there are 1000 cm3 in 1 L, 1.0 cm3 is equal to 0.001 L.
Now, let's calculate the number of moles:
n = (1.0 × 10-11 Pa) × (0.001 L) / (8.314 J/(mol·K) × 293 K)
Solving this equation will give us the number of moles of gas remaining in the container.
Sure, I can help you with part (b).
To determine the number of moles of gas remaining in the container, we can use the ideal gas law:
PV = nRT
Where:
P = pressure (in Pa)
V = volume (in m³)
n = number of moles
R = ideal gas constant (8.314 J/(mol•K))
T = temperature (in Kelvin)
In this case, the volume is still 1.0 cm³, which is equal to 1.0 × 10⁻⁶ m³ (since 1 cm³ = 1 × 10⁻⁶ m³).
The pressure is given as 1.0 × 10⁻¹¹ Pa, and the temperature remains constant.
Now, let's rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the values:
n = (1.0 × 10⁻¹¹ Pa) × (1.0 × 10⁻⁶ m³) / [(8.314 J/(mol•K)) × (20 + 273.15) K]
Calculating the expression:
n ≈ (1.0 × 10⁻¹¹) × (1.0 × 10⁻⁶) / [(8.314) × (293.15)]
n ≈ 4.023 × 10⁻²⁴ mol
So, there are approximately 4.023 × 10⁻²⁴ moles of gas remaining in the container.
Certainly! I can help you with part (b) of your question.
To determine the number of moles of gas remaining in the container when the pressure is reduced to 1.0 x 10^(-11) Pa, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in Pascal)
V = volume (in cubic meters)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
Since the volume is given in cm^3, we need to convert it to cubic meters:
1 cm^3 = 1 x 10^(-6) m^3
Now, let's plug in the known values and solve for the number of moles:
Pressure (P) = 1.0 x 10^(-11) Pa
Volume (V) = 1 x 10^(-6) m^3 (converted from 1.0 cm^3)
R = 8.314 J/(mol·K) (the value of the ideal gas constant)
T = 20 + 273 = 293 K (conversion from Celsius to Kelvin, as temperature must be in Kelvin)
Using the equation PV = nRT, we can rearrange it to solve for n:
n = PV / (RT)
Substituting the values, we have:
n = (1.0 x 10^(-11) * 1 x 10^(-6)) / (8.314 * 293)
Calculating this expression:
n = 1 x 10^(-17) / 2440.002
n ≈ 4.09 x 10^(-21) moles
Therefore, the number of moles of gas remaining in the container is approximately 4.09 x 10^(-21) moles.
I hope this helps you understand how to solve part (b) of your question! Let me know if you have any further queries.
Atmospheric pressure is 1.015*10^5 Pa
A decrease in pressure to 10^-11 Pa is a factor of 10^16 reduction.
At the same temperature, the number of moles and molecules present in the fixed volume must decrease by a factor of 10^16.
The number of moles in part A is the number of molecules divided by Avogadro's number.