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4 years ago

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A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the

gas molecules (only a fourth remain). How must the temperature be changed (as a multiple of T1) to keep the pressure and the volume the same?

1 answer:

cupoosta [38]4 years ago

6 0

Answer:

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

Explanation:

Pressure in the container is P and volume is V.

Temperature of the helium gas molecules = $T_1$

Molecules helium gas = x

Moles of helium has = $n_1= \frac{x}{N_A}$

PV = nRT (Ideal gas equation)

$PV=n_1RT_1$ ...[1]

After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.

Molecules of helium left after removal = $\frac{x}{4}$

Moles of helium has left after removal = $n_2= \frac{x}{4\times N_A}$

$PV=n_2RT_2$ ...[2]

$n_1RT_1=n_2RT_2$

$\frac{x}{N_A}\times T_1=\frac{x}{4\times N_A}\times T_2$

$T_1=\frac{T_2}{4}$

$T_2=4T_1$

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

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hjlf

Answer:

Freezing point is -2.81°C

Explanation:

34g/342gmol^-1 = 0.0994mol

n = m/mr

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3 years ago

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Answer:

Corrosion

Explanation:

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4 years ago

Read 2 more answers

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

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3 years ago