A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure is 880 kPa? (Assume constant volume).

Respuesta :

Answer:

The new pressure at constant volume is 1066.56 kPa

Explanation:

Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.

Mathematically, P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

At constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.

What is Gay-Lussac's law?

Gay-Lussac's law states that the pressure exerted by a given quantity of gas varies directly with the absolute temperature of the gas.

It is expressed as;

P₁/T₁ = P₂/T₂

Given the data in the question;

  • Initial pressure P₁ = 880kPa = ( 880 / 101.325)atm = 8.68492atm
  • Initial temperature T₁ = 250K
  • Initial temperature T₂ = 303K
  • Final pressure P₂ = ?

P₁/T₁ = P₂/T₂

P₁T₂ = P₂T₁

P₂ = P₁T₂ / T₁

P₂ = (8.68492atm × 303K) / 250K

P₂ = (2631.53atmK) / 250K

P₂ = 10.526atm

P₂ = (10.526 × 101.325)kPa

P₂ = 1066.5kPa

Therefore, at constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.

Learn more about Gay-Lussac's law here: brainly.com/question/1358307

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