A compressed gas cylinder contains 1.00 103 g of argon gas. The pressure inside the cylinder is 2096 psi (pounds per square inch) at a temperature of 19°C. How much gas remains in the cylinder if the pressure is decreased to 656 psi at a temperature of 27°C?

Well, let's do some calculations! But before we get started, just remember that I'm here to make you smile, not to bore you with equations.

Now, to solve this problem, we can apply the ideal gas law: PV = nRT. Where P is the pressure, V is the volume, n is the amount of gas in moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's convert the temperatures to Kelvin. To do that, we just need to add 273 to the Celsius temperatures: 19°C + 273 = 292K and 27°C + 273 = 300K.

Now, let's find the amount of gas using the initial conditions. We can rearrange the ideal gas law to solve for n: n = (PV) / (RT).

Using the initial pressure, volume, and temperature:

n1 = (2096 psi * V) / (R * 292K).

Similarly, using the final pressure, volume, and temperature:

n2 = (656 psi * V) / (R * 300K).

Since we are dealing with the same cylinder, we can assume the volume, V, remains constant. Therefore, we can equate n1 and n2:

(2096 psi * V) / (R * 292K) = (656 psi * V) / (R * 300K).

Now, we can cancel out the volume, V, and the ideal gas constant, R, from both sides of the equation. No harm done!

(2096 psi / 292K) = (656 psi / 300K).

Now, let's solve for the ratio:

(2096 psi / 292K) = X / (300K).

Cross-multiply:

(2096 psi * 300K) = (292K * X).

Divide both sides by 292K:

X = (2096 psi * 300K) / 292K.

And after doing the math, we find X = 2150 psi. Therefore, there are 2150 pounds of humor gas remaining in the cylinder! Isn't it funny how that worked out?

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

PV = nRT

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

First, let's convert the given temperature of 19°C to Kelvin:
T1 = 19°C + 273.15 = 292.15 K

Next, let's convert the given pressure of 2096 psi to atm:
P1 = 2096 psi / 14.7 psi/atm = 142.31 atm

We now have P1, V1, and T1, so we can calculate the number of moles of argon gas initially in the cylinder:
PV = nRT
n = PV / RT
n1 = (142.31 atm) * (V1 in L) / (0.0821 L·atm/mol·K * 292.15 K)

Given the mass of argon gas is 1.00 * 10^3 g, we can calculate the molar mass of argon (39.95 g/mol) to find the number of moles initially:
n1 = (1.00 * 10^3 g) / (39.95 g/mol)

Now, let's convert the given temperature of 27°C to Kelvin:
T2 = 27°C + 273.15 = 300.15 K

Next, let's convert the given pressure of 656 psi to atm:
P2 = 656 psi / 14.7 psi/atm = 44.63 atm

We now have P2, V1, and T2, so we can calculate the volume of the gas using the ideal gas law equation:
PV = nRT
V2 = (n1 * R * T2) / P2

Finally, let's calculate the number of moles of argon gas remaining in the cylinder at the new pressure:
n2 = P2 * V2 / (R * T2)

Now, we can calculate the mass of argon gas remaining using the molar mass of argon:
mass remaining = n2 * molar mass of argon

Therefore, the amount of gas remaining in the cylinder can be calculated using these steps.

To find out how much gas remains in the cylinder, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in this case, in psi)
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature (in this case, in Kelvin)

First, we need to convert the temperatures from Celsius to Kelvin.

Temperature in Kelvin = Temperature in Celsius + 273.15

So, 19°C = 19 + 273.15 = 292.15 K
And 27°C = 27 + 273.15 = 300.15 K

Now, let's calculate the number of moles of gas originally in the cylinder:

n1 = (mass of gas / molar mass of gas)
= (1.00 * 10^3 g / 39.948 g/mol)
≈ 25.03 mol

Next, we can use the ideal gas law equation to calculate the volume of gas in the cylinder at the initial conditions:

V1 = (n1 * R * T1) / P1

Where:
V1 = volume at initial conditions
R = ideal gas constant (0.0821 L⋅atm/(mol⋅K) or 8.314 J/(mol⋅K))
T1 = initial temperature in Kelvin
P1 = initial pressure in atmospheres

Let's convert the pressure from psi to atmospheres:

1 atmosphere = 14.7 psi

P1 = (2096 psi / 14.7 psi/atm)
≈ 142 atm

Now we can calculate V1:

V1 = (25.03 mol * 0.0821 L⋅atm/(mol⋅K) * 292.15 K) / 142 atm
≈ 41.07 L

Now we can use the ideal gas law equation again to calculate the number of moles of gas remaining in the cylinder:

n2 = (P2 * V2) / (R * T2)

Where:
n2 = number of moles of gas remaining
P2 = final pressure in atmospheres
V2 = final volume
T2 = final temperature in Kelvin

Let's convert the final pressure from psi to atmospheres:

P2 = (656 psi / 14.7 psi/atm)
= 44.63 atm

Now, we need to find the final volume. Since the number of moles of gas remaining should be the same as before, we can use the equation:

n1 = n2

(25.03 mol * 0.0821 L⋅atm/(mol⋅K) * 292.15 K) / 142 atm = (44.63 atm * V2) / (0.0821 L⋅atm/(mol⋅K) * 300.15 K)

Simplifying the equation, we find:

V2 ≈ 29.83 L

Therefore, approximately 29.83 L of gas remain in the cylinder if the pressure is decreased to 656 psi at a temperature of 27°C.

Convert 2097 psi to atm.

2097 psi/14.7 = ? atm
Convert 1000 g to mol Ar. mol = grams/molar mass.

Use PV = nRT, substitute and solve for V = volume.

Then reuse PV = nRT, substitute the new P, the old V, R, and new T and solve for n. Then n = grams/molar mass, you know n and molar mass, solve for grams.
Don't forget T must be in kelvin.