Question

A 3.0 L flask containing helium at 145 mmHg is connected by a closed valve to a 2.0 L flask containing argon at 355 mmHg. When the valve is opened and the gases are allowed to mix equally in the two flasks, what is the total pressure (in mmHg) in the two connected flasks after mixing ?

Answer 1

The total pressure is 229 atm

Step-by-step explanation:

Step 1: Data given

Volume of helium flask = 3.0 L

Pressure helium flask = 145 mm Hg

Volume of argon flask = 2.0 L

Pressure argon flask = 355 mmHg

total volume = 5.0 L

Step 2: Partial pressure helium

pHe = 145 *(3/5) = 87.0 atm

Step 3: Calculate pressure argon

pAr = 355*(2/5) = 142.0 atm

Step 4: Calculate total pressure

Total pressure = 87.0 + 142.0 atm

Total pressure = 229 atm

The total pressure is 229 atm

Answer 2

Assuming that both helium and argon act like ideal gases, the total pressure after mixing would be approximately
`229\; \rm mmHg`.

Step-by-step explanation:

By the ideal gas equation,
`P\cdot V = n \cdot R \cdot T`, where

• 
  `P` is the pressure of the sample.
• 
  `V` is the volume of the container.

• 
  `n` is the number of moles of gas particles in the sample.
• 
  `R` is the ideal gas constant.
• 
  `T` is the temperature of the sample.

Rewrite to obtain:

• 
  `\displaystyle n = (P \cdot V)/(R\cdot T)`, and
• 
  `\displaystyle P = (n \cdot R \cdot T)/(V)`.

Assume that the two samples have the same temperature,
`T`. Also, assume that mixing the two gases did not affect the temperature.

Apply the equation
`\displaystyle n = (P \cdot V)/(R\cdot T)` to find the number of moles of gas particles in each container:

• In the helium container,
  `V = 3.0\; \rm L` and
  `P = \rm 145\; mmHg`. Hence,
  `\displaystyle n_1 = (P\cdot V)/(R \cdot T) = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T}`.
• In the argon container,
  `V = 2.0\; \rm L` and
  `P = 355\; \rm mmHg`. Hence,
  `\displaystyle n_2 = (P\cdot V)/(R \cdot T) = \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}`.

After mixing,
`V = 2.0 + 3.0 = 5.0\; \rm L`. Assuming that temperature
`T` stays the same.

`\displaystyle n_1 + n_2 = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}`.

Apply the equation
`\displaystyle P = (n \cdot R \cdot T)/(V)` to find the pressure after mixing.

`\begin{aligned}P &= \displaystyle \frac{\displaystyle \displaystyle \left(\frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}\right) \cdot R \cdot T}{5.0\; \rm L} \\ &= (3.0\; \rm L * 145\; \rm mmHg + 2.0\; \rm L * 355\; \rm mmHg)/(5.0\; \rm L) \\ &\approx 229\; \rm mmHg\end{aligned}`.