Chemistry, 15.02.2020 10:44, ineedhelpireallydo

15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is the total pressure at 51 °C?

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answered: mandy9386

The total pressure of the gas mixture which were transferred to a tank of 6.25 L at 51 °C is 1291.7 KPa

We'll begin by calculating the number of mole of N₂ and O₂ which was transferred. This can be obtained as follow:

For N₂:

Volume (V) = 15.1 L

Temperature (T) = 25 °C = 25 + 273 = 298 K

Pressure (P) = 125 KPa

Gas constant (R) = 8.314 L.KPa/Kmol

Number of mole (n) =?

PV = nRT

125 × 15.1 = n × 8.314 × 298

1887.5 = n × 2477.572

Divide both side by 2477.572

n = 1887.5 / 2477.572

n = 0.762 mole For O₂:

Volume (V) = 44.3 L

Temperature (T) = 25 °C = 25 + 273 = 298 K

Pressure (P) = 125 KPa

Gas constant (R) = 8.314 L.KPa/Kmol

Number of mole (n) =?

PV = nRT

125 × 44.3 = n × 8.314 × 298

5537.5 = n × 2477.572

Divide both side by 2477.572

n = 5537.5 / 2477.572

n = 2.235 molesNext, we shall determine the total mole of the mixture.

Mole of N₂ = 0.762 mole

Mole of O₂ = 2.235 moles

Total mole = 0.762 + 2.235

Total mole = 2.997 moles Finally, we shall determine the total pressure at 51 °C. This can be obtained as follow:

Volume (V) = 6.25 L

Temperature (T) = 51 °C = 51 + 273 = 324 K

Gas constant (R) = 8.314 L.KPa/Kmol

Total of mole (n) = 2.997 moles

Total Pressure (P) =?

PV = nRT

P × 6.25 = 2.997 × 8.314 × 324

P × 6.25 = 8073.126792

Divide both side by 6.25

P = 8073.126792 / 6.25

P = 1291.7 KPa

Therefore, the total pressure of the gas mixture at 51 °C is 1291.7 KPa

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answered: pandaho

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

Explanation:

For N2,

Pressure(P₁)=125 kPa

Volume(V₁)=15·1 L

Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

Then, for the mixture,

Volumeof the mixture( V)= 6.25 L

Pressure(P)=?

Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

or, $6.34+18.58=\frac{P*6.25}{324}$

or, $P=\frac{24.92*324}{6.25}$

∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

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15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a...