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A rigid container holds a gas at a pressure of 55 kPa and a temperature of -100.0*C. What will the pressure be when the temperature is increased to 200*C​

Respuesta :

Answer:

P2 = 150.4 KPa

Explanation:

Gay-Lussac's Law:

at constant volume, the pressure of a gas varies directly with the temperature

  • P1/T1 = P2/T2

∴ P1 = 55 KPa

∴ T1 = - 100.0°C ≅ 173 K

∴ T2 = 200°C ≅ 473 K

⇒ P2 = ?

  • P2 = T2*P1 / T1

⇒ P2 = (473 K)(55 KPa) / (173 K)

⇒ P2 = 150.4 KPa

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