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baherus [9]
3 years ago
14

A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

5 mmHg ; Ar, 119 mmHg ; and O2, 163 mmHg . The mixture also contains helium gas. Part A What is the partial pressure of the helium gas? PHe P H e = nothing mmHg Request Answer Part B What mass of helium gas is present in a 10.2-L sample of this mixture at 283 K ? m m = nothing g Request Answer
Chemistry
1 answer:
Setler79 [48]3 years ago
6 0

<u>Answer:</u>

<u>For Part A:</u> The partial pressure of Helium is 218 mmHg.

<u>For Part B:</u> The mass of helium gas is 0.504 g.

<u>Explanation:</u>

  • <u>For Part A:</u>

We are given:

p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg

To calculate the partial pressure of helium, we use the formula:

P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}

Putting values in above equation, we get:

745=245+119+163+p_{He}\\p_{He}=218mmHg

Hence, the partial pressure of Helium is 218 mmHg.

  • <u>For Part B:</u>

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

PV=\frac{m}{M}RT

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g

Hence, the mass of helium gas is 0.504 g.

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