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a sample of nitrogen gas is stored in a 500.0 mL flask at 108 kPa and 10.0C. the gas is transferred to a 750.0 mL flask at 21.0C. What is the pressure of nitrogen in the second flask?

Respuesta :

we can use the combined gas law equation to find the new pressure in the second flask 
[tex] \frac{P1V1}{T1} = \frac{P2V2}{T2} [/tex]
where 
P - pressure
V - volume 
T - temperature in kelvin
T1 - 10.0 °C + 273 = 283 K
T2 - 21.0 °C + 273 = 294 K
substituting the values in the equation 
[tex] \frac{108 kPa *500.0 mL}{283 K} = \frac{P*750.0 mL }{294 K} [/tex]
 P = 74.8 kPa
new pressure is 74.8 kPa 
Eduard22sly

The pressure of nitrogen gas in the second flask given the data is 74.8 KPa

Data obtained from the question

  • Initial volume (V₁) = 500 mL
  • Initial pressure (P₁) = 108 KPa
  • Initial temperature (T₁) = 10° C = 10 + 273 = 283 K
  • New Volume (V₂) = 750 mL
  • New temperature (T₂) = 21 °C = 21 + 273 = 294 K
  • New pressure (P₂) = ?

How to determine the new pressure

The new pressure of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

(108 × 500) / 283 = (P₂ × 750) / 294

Cross multiply

P₂ × 750 × 283 = 108 × 500 × 294

Divide both side by 750 × 283

P₂ = (108 × 500 × 294) / (750 × 283)

P₂ = 74.8 KPa

Learn more about gas laws:

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