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3 years ago

12

a heliox tank contains 32% helium and 68% oxygen. the total pressure in the tank is 395 kPa. What is the partial pressure of oxy

gen in the tank?

1 answer:

trapecia [35]3 years ago

6 0

Answer:

Pp O2 = 82.944 KPa

Explanation:

heliox tank:

∴ %wt He = 32%

∴ %wt O2 = 68%

∴ Pt = 395 KPa

⇒ Pp O2 = ?

assuming a mix of ideal gases at the temperature and volumen of the mix:

∴ Pi = RTni/V

∴ Pt = RTnt/V

⇒ Pi/Pt = ni/nt = Xi

⇒ Pi = (Xi)*(Pt)

∴ Xi: molar fraction (ni/nt)

⇒ 0.68 = mass O2/mass mix

assuming mass mix = 100 g

⇒ mass O2 = 68 g

∴ molar mass O2 = 32 g/mol

⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2

⇒ mass He = 32 g

∴ molar mass He = 4.0026 g/mol

⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He

⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles

molar fraction O2:

⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099

⇒ Pp O2 = (X O2)(Pt)

⇒ Pp O2 = (0.2099)(395 KPa)

⇒ Pp O2 = 82.944 KPa

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<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

<u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

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q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

<u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

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$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

<u>For b:</u>

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

<u>For process 2:</u>

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

<u>For process 3:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

<u>For process 4:</u>

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

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