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Sonja [21]
2 years ago
3

A gas storage tank has a volume of 3.5 x 10^5 m^3 when the temperature is 27 C and the pressure is 101 kPa. What os the new volu

me of the tank if the temperature drops to -10 C and the pressure drops to 95 kPa
Chemistry
1 answer:
victus00 [196]2 years ago
3 0

Answer:

3.26×10⁵m³

Explanation:

Given data:

Initial volume = 3.5×10⁵ m³

Initial temperature = 27 °C

Initial pressure = 101 Kpa

Final temperature = -10°C

Final pressure = 95 Kpa

Final volume = ?

Solution:

Initial temperature = 27 °C (27+273= 300 K)

Final temperature = -10°C (-10+273 = 263 K)

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 101 Kpa × 3.5×10⁵ m³× 263 K / 300 K× 95 Kpa  

V₂ = 92970.5 ×10⁵ kpa.m³. K / 28500 K. Kpa

V₂ = 3.26×10⁵m³

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Explanation : Given,

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Volume of NaHCO_3 = V_2 = 30 mL = 0.03 L

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\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol

and,

\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol

Now we have to calculate the moles of Na^+ ions.

As, 1 mole of Na_2CO_3 will give 2 moles of Na^+ ions

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As, 1 mole of NaHCO_3 will give 1 mole of Na^+ ions

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Total number of moles of Na^+ ions = 0.42 + 0.03 =0.45 mole

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Now we have to calculate the concentration of Na^+ ions.

\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M

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2 years ago
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P1 = ?

P2 = 2ATM

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P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

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P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

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2 years ago
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3 0
2 years ago
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