My teacher gave us a packet with many questiona on it for homework. I was able to figure all 44 questions except the following 2.... Help!?!?!
9. "In order to have 1.00 mole of gas fit in a box that measures 1.30 dm x 2.40 dm x 5.83 dm at 1.00 atm, what must the temperature be (in C)? (1L=1dm^3)"
10. "A cube shaped box is to be made that can hold precisely 40.0 grams of He at 1.05 atm and 55C. How long would the box have to be? (remember it's a cube so take the cube root of the formula) "
at STP 1 mol is 22.4 liters
n = P V/R T
1 = 1*22.4 /R(273)
R = 22.4/273
Now our problem
1 = 1 * 18.2 *273/ (22.4)K
K = 222
222-273 = - 51 deg C
55+ 273 = 328 K
He = 4 g/mol so we have 10 moles
P1V1/(n1 T1) = P2V2/(n2 T2)
22.4/273 = 1.05 V2/(10*328)
V2 = 269 dm^3
side of cube = 6.48 dm
= 64.8 cm or 0.648 m
9. Well, to fit 1.00 mole of gas in a box, you might need a magic wand or a shrink ray! Or maybe you could ask the gas nicely to squish itself. Just kidding!
To solve this problem, we can use the ideal gas law equation: PV = nRT. Since we have the pressure (1.00 atm), volume (1.30 dm x 2.40 dm x 5.83 dm = ??), and moles (1.00 mole), we just need to find the temperature (in C).
Now, let's go back to the equation. We need to convert the volume from dm^3 to liters, so divide by 1,000. Plug in the values and solve for T.
But remember, laughter is the best way to stay warm! So, here's a joke to keep you smiling as you calculate: Why did the scarecrow win an award? Because he was outstanding in his field!
10. Ah, a cube-shaped box! Geometry and gas - the perfect combination. To calculate the length of the box, we need to find the volume first.
But wait, don't panic! We know the mass of helium (40.0 grams), the pressure (1.05 atm), and the temperature (55C). Convert the temperature to Kelvin by adding 273.15.
Now, we can use the ideal gas law equation again. Remember, the volume of a cube is just side^3. Find the volume using the given values and solve for the length.
And here's a funny thought to keep you entertained: Why don't scientists trust atoms? Because they make up everything!
9. To solve this question, we can use 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 gas constant (0.0821 L·atm/(mol·K)), and T is the temperature (in Kelvin).
Given:
P = 1.00 atm
V = 1.30 dm x 2.40 dm x 5.83 dm = 18.420 dm³ (converted to liters)
n = 1.00 mole
R = 0.0821 L·atm/(mol·K)
Plug these values into the ideal gas law equation and solve for T:
1.00 atm x 18.420 L = 1.00 mol x 0.0821 L·atm/(mol·K) x T
Simplifying the equation:
18.420 atm·L = 0.0821 L·atm/(mol·K) x T
Divide both sides of the equation by (0.0821 L·atm/(mol·K)):
T = (18.420 atm·L)/(0.0821 L·atm/(mol·K))
Evaluating the expression:
T ≈ 225 K
To convert from Kelvin to Celsius, you need to subtract 273.15 from the temperature in Kelvin:
T ≈ 225 K - 273.15 = -48.15°C
Therefore, the temperature required for 1.00 mole of gas to fit in the given box is approximately -48.15°C.
10. To solve this question, we can use the ideal gas law equation (PV = nRT) again.
Given:
P = 1.05 atm
n = 40.0 g / molar mass of He
R = 0.0821 L·atm/(mol·K)
T = 55°C + 273.15 = 328.15 K
First, we need to find the number of moles of He:
molar mass of He = 4.00 g/mol
n = 40.0 g / 4.00 g/mol
n = 10.0 mol
Now, let's solve for volume (V) using the ideal gas law equation:
P x V = n x R x T
1.05 atm x V = 10.0 mol x 0.0821 L·atm/(mol·K) x 328.15 K
Simplifying the equation:
1.05 atm x V = 269.073 L·atm
To find the volume (V), divide both sides of the equation by 1.05 atm:
V ≈ 269.073 L·atm / 1.05 atm
V ≈ 256.45 L
Since the box is a cube, all sides have the same length. To find the length (L), we need to calculate the cube root of the volume:
L = ∛(256.45 L)
Using a calculator, ∛(256.45) ≈ 6.34 L
Therefore, the cube-shaped box needs to have a length of approximately 6.34 L in order to hold precisely 40.0 grams of Helium at 1.05 atm and 55°C.
Sure, I'd be happy to help you with those questions!
For question 9, you need to use the Ideal Gas Law equation, which is given by:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L∙atm∙mol^−1∙K^−1)
T = temperature (in Kelvin)
To solve the problem, we need to convert the given volume from dm^3 to liters since the unit of liters is used in the Ideal Gas Law equation. We are given that 1L = 1dm^3, so the given volume of the box is 1.30 dm x 2.40 dm x 5.83 dm = 18.0588 dm^3.
Now, we need to convert this volume to liters by dividing it by 1L/1dm^3, which gives us 18.0588 L of volume.
Additionally, we are given that the pressure (P) is 1.00 atm, and we want to find the temperature (T) in Celsius (°C).
Substituting the known values into the Ideal Gas Law equation, we get:
(1.00 atm) * (18.0588 L) = (1 mole) * (0.0821 L∙atm∙mol^−1∙K^−1) * (T)
Simplifying this equation, we can cancel out the units:
18.0588 L∙atm = 0.0821 L∙atm∙K^-1 * T
Now, we can solve for T by rearranging the equation:
T = (18.0588 L * atm) / (0.0821 L∙atm∙K^-1)
Calculating this value will give you the temperature in Kelvin (K). To convert it to Celsius (°C), you need to subtract 273.15 from the obtained value.
For question 10, we can use the combined gas law equation to solve for the length of the cube box.
The combined gas law equation is given by:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Where:
P₁ and P₂ = initial and final pressures (in atm)
V₁ and V₂ = initial and final volumes (in liters)
T₁ and T₂ = initial and final temperatures (in Kelvin)
We are given the values for pressure (P₂ = 1.05 atm), temperature (T₂ = 55°C = 55 + 273.15 = 328.15 K), and we need to find the length (L) of the cube-shaped box.
Let's assume that the length of each side of the cube is L. Since it's a cube, the volume (V₂) of the box is given by L³.
Substituting the known values in the combined gas law equation, we get:
(1.05 atm * V₁) / (55 + 273.15 K) = (1 atm * L³) / 328.15 K
Simplifying this equation gives us:
(1.05 atm * V₁) / 328.15 K = L³ / 328.15 K
Now, we can solve for L by taking the cube root of both sides:
L = (1.05 atm * V₁)^(1/3)
Calculate this value by substituting the given volume of 40.0 grams of He into the equation and taking the cube root. Remember to convert the mass of He to moles by dividing it by the molar mass of He.
I hope this explanation helps you solve those questions! Let me know if you have any further questions.