After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

Here's how I go about doing this.

What we can do is use the ideal gas equation for both $"Ar"$ and $"O"_2$ to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume ($400$ $"mL"$) and temperature ($24^"o""C"$).

I won't show the here, as I predict you already know how; the moles of each substance is

$n_ "Ar" = ((1.47cancel("atm"))(0.600cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(494cancel("K"))) = 0.0218$ $"mol Ar"$

$n_ ("O"_2) = ((0.580cancel("atm"))(0.200cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(388cancel("K"))) = 0.00365$ $"mol O"_2$

$n_"total" = 0.0218$ $"mol Ar"$ $+ 0.00365$ $"mol O"_2$ $= 0.0254$ $"mol"$

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

$P = (nRT)/V = ((0.0254cancel("mol"))(0.082057(cancel("L")•"atm")/(cancel("mol")•cancel("K")))(297cancel("K")))/(0.400cancel("L"))$

$= color(blue)(1.55$ $color(blue)("atm"$

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
What is the total pressure of a container that has $NH_3(g)$ exerting a pressure of 34 torr, $N_2(g)$ exerting a pressure of 225 torr, and $H_2O(g)$ exerting a pressure of 55 torr?

According to Dalton's law of partial pressures, $P_(t otal) = P_1 + P_2 + P_3 ...$ So the total pressure of the system is merely the partial pressures of each...

1 Answers 1 views
After 0.600 L of $Ar$ at 1.40 atm and 210 $"^o$C is mixed with 0.200L of $O_2$ at 398 torr and 120 $"^o$C in a 400 -mL flask at 29 $"^o$C, what is the pressure in the flask?

$n_"Ar"=(1.40*atmxx0.600*L)/(0.0821*(L*atm)/(K*mol)xx483.15*K)=0.0212*mol$ $n_(O_2)=((398*mm)/(760*mm*Hg*atm^-1)xx0.200*L)/(0.0821*(L*atm)/(K*mol)xx393.15*K)=0.00325*mol$ And now we ADD these molar quantities, and calculate the pressure expressed under the NEW given conditions... $P_"mixture"=(0.245*molxx0.0821*(L*atm)/(K*mol)xx302.15*K)/(400*mLxx10^-3*L*mL^-1)$ I make this approx. $15*atm$ but do not trust my...

1 Answers 1 views
The total pressure of a container that has $NH_3(g)$ exerting a pressure of 346 torr, $N_2(g)$ exerting a pressure of 225 torr, and $H_2O (g)$ exerting a pressure of 55 torr?

And thus....................... $P_"total"=P_"ammonia"+P_"dinitrogen"+P_"water vapour"$ $=(346+225+55)$ $mm*Hg$ $=626*"Torr"$. How many atmospheres is this? Can you think of a way to estimate the temperature at which this experiment was performed? I can...

1 Answers 1 views
A mixture of neon and oxygen gases, in a 9.77 L flask at 65 °C, contains 2.84 grams of neon and 7.67 grams of oxygen. The partial pressure of oxygen in the flask is ? atm and the total pressure in the flask is atm?

We can use the to calculate the partial pressures. $color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "$ or $p = (nRT)/V$ The number of moles $n$...

1 Answers 1 views
After 0.600 L of Ar at 1.29 atm and 223 degrees Celsius is mixed with 0.200 L of O2 at 535 torr and 106 degrees Celsius in a 400.-mL flask at 23 degrees Celsius, what is the pressure in the flask?

Concept of is to be applied here. Let the partial pressure of $Ar$ in the final mixture be $p_"Ar"$ Initially before mixing $ Ar$ had Pressure $P_"Ar"=1.29$ atm Temperature$T_"Ar"=223+273=500$...

1 Answers 1 views
A sample of argon gas is collected over water at 30.0 degrees C. The level of the water is adjusted until the total pressure in the flask is 771 torr and the volume is 250.0 mL. The vapor pressure of water is 31.8 torr at 30.0 degrees C.?

Partial pressures are additive. The pressure exerted inside the flask is that due to the argon AND the water vapour. For $n_(Ar)$, we simply solve the Ideal Gas Equation: $n=(PV)/(RT)$...

1 Answers 1 views
A flask contains oxygen, carbon dioxide, and nitrogen gases. The partial pressures of each are 11.4 atm, 5.6 atm, and 8.7 atm respectively. What is the total pressure in the flask in atm?

The total pressure is the sum of the pressures of the three gases in the flask. $"pressure"_"total" ="11.4 atm + 5.6 atm + 8.7 atm" = "25.7 atm"$

1 Answers 1 views
A flask contains helium at $752*"Torr"$, and a second flask contains argon at $712*"Torr"$. What are the partial pressures of each gas in each flask?

In a gaseous mixture, the partial pressure exerted by component is the same as the pressure it would exert if it alone occupied the container. The total pressure is the...

1 Answers 1 views
A mixture of $"N"_2$, $"O"_2$. and $"CO"_2$ has a total pressure of $"740 torr"$ In this mixture, the partial pressure of $"N"_2$ is $"120 torr"$ and the partial pressure of $"O"_2$ is $"400 torr"$. What is the partial pressure of $"CO"_2$ ?

From Dalton's Law of , the total pressure of a mixture of gases is the sum of the partial pressures of each gas. Therefore, the partial pressure of $"CO"_2$ is...

1 Answers 1 views
A sample of carbon is placed in a rigid 1.50 L flask. Dinitrogen monoxide is added and the flask heated to 800C. When the system reached equilibrium the partial pressure of the carbon dioxide is found to be 0.030 atm and the partial pressure of the dinitrogen monoxide to be 1.48 atm. What is the partial pressure of nitrogen in the mixture.

The of nitrogen is 0.060 atm. First, write the balanced chemical equation for the equilibrium. C(s) +2N₂O(g) ⇌ CO₂(g) + 2N₂(g) There was no CO₂ or N₂ present at the...

1 Answers 1 views
X