A 5.0x10^-3 m^3 container holds 4.90 g gas when the pressure is 75.0 cm Hg and the temperature is 50.0 degrees C. What will be the pressure if 6.00 g of this gas is confined to a 2.0x10^-3 m^3 container at 0.0 degrees C?
Do I simply need to use the combined gas law or does the fact that they gave the masses change the rules? Thank you very much!!
the mass is proportional to the number of mols n2/n1 = m2/m1
P1 V1 = n1 R T1 so n1 = P1 V1/R T1
P2 V2 = n2 R T2 so n2 = P2 V2/R T2
n2/n1 = m2/m1 = (P2 V2/P1V1)(T1/T2)
6/4.9 = (P2*2/75*5)(323/273)
solve for P2
Well, if you're looking for a straightforward answer, then yes, you can use the combined gas law to solve this problem. However, if you're looking for a more complicated and clown-like answer, let's see what we can come up with!
First, we need to convert the temperature from Celsius to Kelvin because Kelvin is just cooler. So, 50.0 degrees C + 273.15 = 323.15 K.
Now, let's calculate the initial pressure using the ideal gas law equation, PV = nRT. We know the volume (5.0x10^-3 m^3), mass (4.90 g), R (the ideal gas constant), and the temperature (323.15 K), but we don't know the number of moles (n) yet. The number of moles can be calculated by dividing the mass of the gas by its molar mass.
Assuming we know the molar mass of this mysterious gas (and that it's not just hot air), we can calculate the number of moles. But since it's a secret, let's just pretend the gas is made up of magical clown atoms and its molar mass is 123 g/mol.
Calculating the number of moles (n) = mass / molar mass, n = 4.90 g / 123 g/mol = 0.0398 mol.
Now, using the ideal gas law equation, PV = nRT, we can solve for the initial pressure (P).
P * 5.0x10^-3 m^3 = 0.0398 mol * R * 323.15 K
P = (0.0398 mol * R * 323.15 K) / 5.0x10^-3 m^3
Now, I could tell you the exact value of P, but what's the fun in that? Instead, let's just say the initial pressure is equivalent to running away from a hungry lion while juggling flaming torches – it's pretty intense!
Next, we need to find the new pressure when the gas is confined to a smaller container. We can use the combined gas law, which states that P1 * V1 / T1 = P2 * V2 / T2.
Now, we know the initial pressure (P1) and temperature (T1), and we need to find the new pressure (P2) when the volume changes from 5.0x10^-3 m^3 to 2.0x10^-3 m^3, and the temperature changes from 323.15 K to 0.0 degrees C, which we'll convert to 273.15 K because clowns are all about conversions!
Using the combined gas law:
(P1 * V1) / T1 = (P2 * V2) / T2
(P1 * 5.0x10^-3 m^3) / 323.15 K = (P2 * 2.0x10^-3 m^3) / 273.15 K
Now, let's swap the numbers around (like a clown juggling balls) and solve for P2!
P2 = (P1 * 5.0x10^-3 m^3 * 273.15 K) / (323.15 K * 2.0x10^-3 m^3)
And there you have it, the new pressure (P2) with a mix of circus-like calculations and clown magic!
To solve this problem, you will indeed need to use the combined gas law. The fact that they gave the masses does not change how you approach the problem.
The combined gas law relates the initial and final conditions of pressure (P), volume (V), and temperature (T) of a gas sample:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure
V2 = Final volume
T2 = Final temperature
Given:
P1 = 75.0 cm Hg = 75.0 * 1.33322 Pa (since 1 atm = 101325 Pa)
V1 = 5.0x10^-3 m^3
T1 = 50.0 degrees C = 50.0 + 273.15 K
P2 = ?
V2 = 2.0x10^-3 m^3
T2 = 0.0 degrees C = 0.0 + 273.15 K
We need to convert the units of pressure and temperature to SI units before substituting them into the formula.
P1 = 75.0 * 1.33322 Pa = 99,991.5 Pa
T1 = 50.0 + 273.15 K = 323.15 K
Substituting the given values into the combined gas law equation:
(99,991.5 * 5.0x10^-3) / 323.15 = (P2 * 2.0x10^-3) / (0.0 + 273.15)
Now we can solve for P2:
P2 = [(99,991.5 * 5.0x10^-3) / 323.15] * (0.0 + 273.15) / 2.0x10^-3
P2 = 96,759.36 Pa
Thus, the pressure when 6.00 g of the gas is confined to a 2.0x10^-3 m^3 container at 0.0 degrees C will be approximately 96,759.36 Pa.
To solve this problem, you need to use the combined gas law equation, which relates the initial and final states of a gas sample. The combined gas law equation is given by:
(P1 * V1)/(T1) = (P2 * V2)/(T2)
Where:
P1 and P2 are the pressures of the gas
V1 and V2 are the volumes of the gas
T1 and T2 are the temperatures of the gas
In this case, you are given:
P1 = 75.0 cm Hg
V1 = 5.0x10^-3 m^3
T1 = 50.0 degrees C (which needs to be converted to Kelvin, by adding 273.15)
You are also given the mass of the gas in the first container (4.90 g), but it is not necessary to use this information in the calculation.
To solve for P2, you are asked to find the pressure when 6.00 g of the gas is confined to a different volume (V2 = 2.0x10^-3 m^3) and at a different temperature (T2 = 0.0 degrees C, or 273.15 K).
Now, let's plug in the values into the combined gas law equation and calculate P2:
(P1 * V1)/(T1) = (P2 * V2)/(T2)
Substituting the known values:
(75.0 cm Hg * 5.0x10^-3 m^3)/(50.0 + 273.15 K) = (P2 * 2.0x10^-3 m^3)/(0.0 + 273.15 K)
Simplifying the equation, we can cross-multiply and solve for P2:
(75.0 cm Hg * 5.0x10^-3 m^3 * 273.15 K) = (P2 * 2.0x10^-3 m^3 * (50.0 + 273.15 K))
P2 = (75.0 cm Hg * 5.0x10^-3 m^3 * 273.15 K) / (2.0x10^-3 m^3 * (50.0 + 273.15 K))
By substituting the values into the equation and performing the calculation, you can find the value of P2.