A compressed gas cylinder contains 1.00 103 g of argon gas. The pressure inside the cylinder is 2035 psi (pounds per square inch) at a temperature of 15°C. How much gas remains in the cylinder if the pressure is decreased to 610. psi at a temperature of 30.°C?

Thanks for the help in advance :)

Do you mean that you open the valve on the cylinder and release Ar gas until the pressure is decreased to 610 psi at 30 C?

ohh yes i think so

I would go to google, type in "2035 psi to atmospheres" without the quotation marks. Do the same for 610 psi. Then use PV = nRT to calculate n remaining in the vessel and convert that to grams. Subtract from the initial amount to obtain the grams Ar remaining.

i found both atmospheres

2035=138.473537
610=41.508038

so it would be 1.00103(2035)=?

I don't think so. You haven't used PV = nRT

I would substitute P, V, R, and T (don't forget to change T t Kelvin) from the final conditions and calculate n, number of moles. Then moles = grams/molar mass and determine grams, then subtract from the initial grams to find that remaining.

To calculate the amount of gas remaining in the cylinder, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin

First, let's convert the given temperatures from Celsius to Kelvin:

T1 = 15°C + 273.15 = 288.15 K (initial temperature)
T2 = 30°C + 273.15 = 303.15 K (final temperature)

Next, we need to find the volume of the gas at both initial and final states. However, the given question does not provide the volume, so we assume it remains constant throughout the process. This assumption is reasonable if the cylinder is rigid and the change in pressure is not too large.

Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):

n = (PV) / (RT)

Let's calculate the number of moles of argon gas at the initial state:

P1 = 2035 psi
T1 = 288.15 K

n1 = (P1 * V) / (R * T1)

Similarly, we can calculate the number of moles of argon gas at the final state:

P2 = 610 psi
T2 = 303.15 K

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

As mentioned earlier, the volume (V) remains constant, so it cancels out when calculating the ratio of moles:

n2 / n1 = (P2 * V) / (R * T2) * (R * T1) / (P1 * V)

Since R and V cancel out, the equation becomes:

n2 / n1 = (P2 * T1) / (P1 * T2)

Let's substitute the given values and solve for the ratio of moles:

n2 / n1 = (610 psi * 288.15 K) / (2035 psi * 303.15 K)

Now, we know that the ratio of moles (n2 / n1) will be the same as the ratio of gas remaining, as long as the volume remains constant.

Finally, we can calculate the remaining gas by multiplying the ratio of moles by the initial amount of gas:

Remaining gas = (n2 / n1) * 1.00 x 10^3 g

Let's plug in the values and calculate the remaining gas:

Remaining gas = (610 psi * 288.15 K) / (2035 psi * 303.15 K) * 1.00 x 10^3 g

After performing the calculation, you will get the amount of gas, in grams, that remains in the cylinder when the pressure is decreased to 610 psi at a temperature of 30°C.