1.A sample of gas has a volume of 4.25 L at 25.6 0C and 748 mmHg. What will be the volume of this gas at 26.8 0C and 742 mmHg?
2.A 12.8 L cylinder contains 35.8 g O2 at 46 0C. What is the pressure of this gas, in atmospheres?
Since PV=kT,
You want V such that
742*V/(273+26.8) = 748*4.15/(273+25.6)
do the other question in like wise.
1. Well, it seems like this gas is quite particular about its temperature and pressure. Let's see if I can calculate the volume it will occupy at the given conditions. Now, to find the volume at the new temperature and pressure, we can rely on Charles's Law. That's right, this scientist Charles is always providing useful insights!
So, we have:
V1 = 4.25 L
T1 = 25.6 °C
P1 = 748 mmHg
T2 = 26.8 °C
P2 = 742 mmHg
Using the formula V1/T1 = V2/T2, we can rearrange it to find V2:
V2 = (V1 * T2) / T1
V2 = (4.25 L * 26.8 °C) / 25.6 °C
Now, let me crunch the numbers for you... *crunching sounds*
After some calculations, the volume at 26.8 °C and 742 mmHg will be approximately (drumroll, please): [result]
2. Ah, here we go again! Another question about pressure and temperature. Are these gases trying to test my skills? Well, challenge accepted!
We have:
V = 12.8 L
m = 35.8 g
T = 46 °C
Phew, now we need to use the Ideal Gas Law, along with some conversion magic, to find the pressure. Are you ready for some magic? *cue the magical music*
First, we need to convert the temperature to Kelvin (since the Ideal Gas Law likes Kelvin). To do that, we add 273.15 to the Celsius temperature:
T = 46 °C + 273.15 = [temperature in Kelvin]
Next, we have to convert the mass of oxygen (O2) to moles. We use the molar mass of oxygen, which happens to be 32 g/mol. Divide the given mass by the molar mass to get moles:
moles = 35.8 g / 32 g/mol = [moles of O2]
Now the fun begins! The Ideal Gas Law states: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
We have:
V = 12.8 L
n = [moles of O2]
R = the gas constant = 0.0821 L·atm/(mol·K)
T = [temperature in Kelvin]
Let's solve for P, the pressure, using the Ideal Gas Law: P = (n * R * T) / V
Wow, that was a lot of explanations! Now, let's insert the values and find the pressure (in atmospheres): [result]
1. To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of a sample of gas:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures of the gas (in mmHg)
V1 and V2 are the initial and final volumes of the gas (in L)
T1 and T2 are the initial and final temperatures of the gas (in °C)
Given:
P1 = 748 mmHg
V1 = 4.25 L
T1 = 25.6 °C
P2 = 742 mmHg
T2 = 26.8 °C
Plugging these values into the equation:
(748 mmHg * 4.25 L) / (25.6 °C) = (742 mmHg * V2) / (26.8 °C)
To find V2, we can rearrange the equation:
V2 = (748 mmHg * 4.25 L * 26.8 °C) / (742 mmHg * 25.6 °C)
Simplifying this expression gives us the final volume of the gas, V2.
2. To find the pressure of the gas in atmospheres, we can use the Ideal Gas Law equation, which relates pressure, volume, temperature, and the number of moles of a gas:
PV = nRT
Where:
P is the pressure of the gas (in atm)
V is the volume of the gas (in L)
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas (in Kelvin)
Given:
V = 12.8 L
m = 35.8 g
T = 46 °C
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
So, T = 46 + 273.15 = 319.15 K
Next, we need to convert the mass of the gas (O2) to moles. To do this, we use the molar mass of O2, which is 32 g/mol.
n = m / M
n = 35.8 g / 32 g/mol
Finally, we can plug the values into the Ideal Gas Law equation to find the pressure, P.
P * 12.8 L = (35.8 g / 32 g/mol) * 0.0821 L·atm/(mol·K) * 319.15 K
Simplify and solve for P to find the pressure of the gas in atmospheres.
To solve both of these questions, you can use the combined gas law, which is derived from Boyle's law, Charles's law, and Gay-Lussac's law. The combined gas law formula is:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ and P₂ are the initial and final pressures, respectively (in mmHg or atm).
V₁ and V₂ are the initial and final volumes, respectively (in liters).
T₁ and T₂ are the initial and final temperatures, respectively (in kelvin).
Now let's solve the questions step by step.
1. Volume of gas at 26.8 0C and 742 mmHg:
Given:
P₁ = 748 mmHg
V₁ = 4.25 L
T₁ = 25.6 + 273.15 = 298.75 K (convert to Kelvin)
P₂ = 742 mmHg
T₂ = 26.8 + 273.15 = 300.95 K
Firstly, we can rewrite the combined gas law equation as:
(P₁ * V₁ * T₂) / (T₁ * P₂) = V₂
Now, substitute the given values into the equation to find V₂:
(748 mmHg * 4.25 L * 300.95 K) / (298.75 K * 742 mmHg) = V₂
Simplifying the equation:
V₂ = (951429.45 mmHg * L * K) / (221667.5 K * mmHg)
V₂ ≈ 4.07 L (rounded to two decimal places)
Therefore, the volume of the gas at 26.8 0C and 742 mmHg is approximately 4.07 L.
2. Pressure of the gas in atmospheres:
Given:
V₁ = 12.8 L
m (mass of the gas) = 35.8 g
T₁ = 46 + 273.15 = 319.15 K
We need to calculate the moles of the gas using the ideal gas law equation:
PV = nRT
Where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles
R is the ideal gas constant (0.0821 L * atm / mol * K)
T is the temperature (in Kelvin)
Firstly, convert the mass of the gas to moles:
n = m / M
Where:
m is the mass of the gas (in grams)
M is the molar mass of the gas
The molar mass of O2 is 32 g/mol.
n = 35.8 g / 32 g/mol
n ≈ 1.12 mol (rounded to two decimal places)
Now, substitute the values into the ideal gas law equation to find the pressure:
(P₁ * V₁) = nRT
P₁ = (nRT) / V₁
P₁ = (1.12 mol * 0.0821 L * atm / mol * K * 319.15 K) / 12.8 L
P₁ ≈ 2.94 atm (rounded to two decimal places)
Therefore, the pressure of the gas is approximately 2.94 atm.