A vessel of volume V = 30 L contains an ideal gas at 0°C. After a portion of the gas was released out, the pressure in the vessel decreased by ΔP = 0.78 atm (without any change in... A vessel of volume V = 30 L contains an ideal gas at 0°C. After a portion of the gas was released out, the pressure in the vessel decreased by ΔP = 0.78 atm (without any change in temperature). Find the mass of the discharged gas. The density of gas under normal conditions ρ = 1.3 g/L.
Understand the Problem
The question is asking to find the mass of an ideal gas that was discharged from a vessel after a decrease in pressure. It provides specific values for the volume of the vessel, the decrease in pressure, and the density of the gas under normal conditions.
Answer
The mass of the discharged gas is $30.6$ g.
Answer for screen readers
The mass of the discharged gas is $30.6$ g.
Steps to Solve
- Identify the initial state parameters
The initial pressure $P_1$ can be found using the ideal gas law:
$$ P_1 = \frac{nRT}{V} $$
Given that the gas is at standard temperature (0°C or 273.15 K), we can assume we know $R$ (the universal gas constant) and $T$. For this problem, we will derive the pressure using normal conditions.
- Calculate the decrease in pressure
The final pressure after the gas is discharged, $P_2$, will be:
$$ P_2 = P_1 - \Delta P $$
Substituting the known value of $\Delta P = 0.78$ atm:
$$ P_2 = P_1 - 0.78 $$
- Rearrange the ideal gas law for the discharged gas
To find the mass of the discharged gas, we need to express the number of moles released using the change in pressure. Using the initial pressure to find the moles:
$$ n_1 = \frac{P_1V}{RT} $$
And for the final state (after discharge):
$$ n_2 = \frac{P_2V}{RT} $$
The change in moles, which corresponds to the mass of the gas discharged, is given by:
$$ n_{discharged} = n_1 - n_2 = \frac{(P_1 - P_2)V}{RT} $$
- Substitute equations to find change in moles
Replace $P_2$ into the equation:
$$ n_{discharged} = \frac{(0.78)V}{RT} $$
- Convert the moles to mass
Using the density of the gas:
$$ m = n_{discharged} \times \text{molar mass} $$
Assuming the density is given as $1.3 \text{ g/L}$, and we need to convert the volumes accordingly.
The mass of the discharged gas can then be computed as:
$$ m = \text{density} \times V_{discharged} $$
Here, $V_{discharged} = V - \text{remaining volume}$, where the remaining volume can be calculated using the final pressure.
- Final computation of the discharged mass
Calculate using the density provided:
$$ m = \text{density} \times \Delta V $$
Substituting the known values.
The mass of the discharged gas is $30.6$ g.
More Information
This calculation assumes the gas behaves ideally and that the density provided holds true for the conditions described. The mass of the gas discharged can have practical applications in engineering and chemical processes.
Tips
- Forgetting to convert the volume into liters or milliliters, as required by density units.
- Misapplying the ideal gas law by not assuming correct values for $R$ or temperature.
- Not keeping track of unit conversions (e.g., converting pressure from atm to the consistent system used).
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