A 334-mL cylinder for use in chemistry lectures contains 5.225 g of helium at 23 C. How many grams of helium must be releasedto reduce the pressure to 75 atm assuming ideal gas behavior?

Well, if we want to reduce the pressure to 75 atm, we need to release some helium. And I know what you're thinking - why can't we just use a pin and let it all out in one big whoosh? But unfortunately, it's not that simple.

To solve this, we need to use the ideal gas law equation, which is PV = nRT. P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the volume of the cylinder from mL to liters because the ideal gas constant (R) has units of liters-atm/mol-K. So, 334 mL is equal to 0.334 liters.

Next, we need to convert the temperature from Celsius to Kelvin. To do that, we add 273 to the Celsius temperature. So, 23 C + 273 = 296 K.

Now, let's rearrange the ideal gas law equation to solve for n, the number of moles of helium. We divide both sides of the equation by RT to get n = PV / RT.

Plugging in the values we have:
n = (75 atm * 0.334 L) / (0.0821 L-atm/mol-K * 296 K)

After doing the math, we find that n is approximately 9.091 moles of helium.

Finally, we can solve for the mass of helium using the molar mass of helium, which is 4.0026 g/mol. We multiply the number of moles by the molar mass: 9.091 moles * 4.0026 g/mol ≈ 36.385 g.

So, approximately 36.385 grams of helium must be released to reduce the pressure to 75 atm. Let's just hope we release it slowly and not all at once, or else it'll be like a party balloon gone rogue!

To find the mass of helium that must be released to reduce the pressure to 75 atm, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

Let's convert the given volume to liters and the temperature to Kelvin:

334 mL = 0.334 L
23°C = 23 + 273 = 296 K

Now, let's rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

n = (75 atm) * (0.334 L) / (0.0821 L·atm/mol·K * 296 K)
n = 2.517 mol

Since 1 mole of helium is equal to 4.003 g, we can calculate the mass of helium:

mass = n * molar mass
mass = 2.517 mol * 4.003 g/mol
mass ≈ 10.073 g

Therefore, approximately 10.073 grams of helium must be released to reduce the pressure to 75 atm.

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

First, we need to convert the volume from mL to liters:

334 mL = 334/1000 = 0.334 L

Next, we convert the temperature from Celsius to Kelvin:

T(K) = T(C) + 273.15
T(K) = 23 + 273.15 = 296.15 K

Now, we can plug in the values into the ideal gas law equation and solve for n (number of moles):

PV = nRT
n = PV / RT

Given:
P(initial) = 1 atm
P(final) = 75 atm
V = 0.334 L
R = 0.0821 L·atm/(mol·K)
T = 296.15 K

First, we find the initial number of moles (n_initial):

n_initial = P(initial) * V / (R * T)

n_initial = 1 atm * 0.334 L / (0.0821 L·atm/(mol·K) * 296.15 K)
n_initial ≈ 0.0142 mol

Next, we need to find the final number of moles (n_final) when the pressure is reduced to 75 atm:

n_final = P(final) * V / (R * T)

n_final = 75 atm * 0.334 L / (0.0821 L·atm/(mol·K) * 296.15 K)
n_final ≈ 0.532 mol

To find the number of moles released, we subtract the final number of moles from the initial number of moles:

n_released = n_initial - n_final
n_released ≈ 0.0142 mol - 0.532 mol
n_released ≈ -0.5178 mol

Since we cannot have negative moles, the result is not physically meaningful. Therefore, the pressure cannot be reduced to 75 atm by releasing helium from the cylinder.

find grams at 75

PV=nRT solve for n, then convert to grams.

He is 4 grams/mol

so we have 5.225/4 = 1.31 mols in 0.334 Liter at 273 + 23 = 296 K

P V = n R T

R = 0.082 L atm deg K / mol
V = 0.334 L
n = 1.31 mol
T = 296

so
P = 1.31 * 0.082 * 296 / 0.334
= 95.2 atm
new P = 75
T R V the same
so P is proportional to n
75/95.2 = new n/ 1.31
new n = 1.03 mols
1.03 * 4 g/mol = 4.13 grams remain
5.225 - 4.13 = 1.1 grams released