Two identical gas cylinders each contain 20.0 kg of compresses air at 1.00 x 106 pa and 275 K. One of the cylinders if fitted with a safety valve that releases air into the atmosphere if and only if the pressure in the cylinders raises above 1.10 x106 pa (the other cylinder does not have a pressure release valve). The temperature of the cylinders is then raised to 310 K. Determine: (a) The pressure in the cylinder without the pressure release valve.
(b) The mass of gas that escapes from the cylinder with the pressure release valve.
FIGURE THE MOLES OF GAS ORIGINALLY.
N=pv/rt p,v,t GIVEN
AFTER THE NEW TEMP, FIGURE THE MOLES remaining
n'=Pv/T'
so mass of gas escaped = 27(n-n')
a) is done on the other post.
To solve this problem, we need to apply the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
Let's start with part (a) and find the pressure in the cylinder without the pressure release valve.
Given:
Pressure in the cylinder with the release valve (P1) = 1.00 x 106 Pa
Pressure in the cylinder after raising the temperature (P2) = ? (we need to find this)
Temperature before raising (T1) = 275 K
Temperature after raising (T2) = 310 K
Since we have two identical cylinders, their volumes are the same. Therefore, the V1 and V2 terms cancel out, and we're left with P1/T1 = P2/T2.
Using this equation, we can solve for P2:
P2 = (P1 * T2) / T1
= (1.00 x 106 Pa * 310 K) / 275 K
≈ 1.13 x 106 Pa
So, the pressure in the cylinder without the pressure release valve (P2) is approximately 1.13 x 106 Pa.
Moving on to part (b), we need to find the mass of gas that escapes from the cylinder with the pressure release valve. Since the cylinders are identical, the mass of gas in both cylinders is the same initially.
Given:
Mass of gas in each cylinder (m1 and m2) = 20.0 kg
Pressure in the cylinder with the release valve (P1) = 1.00 x 106 Pa
Pressure after raising the temperature (P2) = 1.13 x 106 Pa
First, we need to calculate the number of moles for each condition using the equation:
n = (m * R) / (M * V)
where m is the mass of gas, R is the ideal gas constant, M is the molar mass of air, and V is the volume (which cancels out since they are identical).
Since the molar mass of air is 29 g/mol and the volume is the same for both cylinders, the moles of gas in each condition can be given as n1 = (m1 * R) / (M * V) and n2 = (m2 * R) / (M * V).
Now, since the mass of the gas is constant, the difference in moles between the two conditions will be the amount of gas that escaped.
Δn = n2 - n1 = [(m2 * R) / (M * V)] - [(m1 * R) / (M * V)]
= [(m2 - m1) * R] / (M * V)
Substituting the given values:
Δn = [(20.0 kg - 20.0 kg) * 8.314 J/(mol·K)] / (29 g/mol * V)
As we already calculated the pressure in part (a), we can use the ideal gas law equation to derive the volume V using the equation PV = nRT.
V = (n2 * R * T2) / P2
= [(m2 * R) / (M * V)] * R * T2 / P2
= [(20.0 kg * 8.314 J/(mol·K)) / (29 g/mol * V)] * 8.314 J/(mol·K) * 310 K / (1.13 x 106 Pa)
Now, since V appears on both sides of the equation, we can solve for V by rearranging the equation and solving iteratively until we converge to the correct value.
After obtaining the value of V, we can substitute it back into the equation for Δn and calculate the mass of gas that escaped from the cylinder.
Δn ≈ [(20.0 kg - 20.0 kg) * 8.314 J/(mol·K)] / (29 g/mol * V)
Finally, we multiply Δn by the molar mass of air (M) to find the mass of gas that escaped.
Mass of gas that escaped = Δn * M
Please note that the equation for volume V requires an iterative solution, and the actual calculations involving substitution and simplification might vary slightly depending on the specific values and units used.