Gas is confined in a tank at a pressure of 9.8 atm and a temperature of 29.0Ā°C. If two-thirds of the gas is withdrawn and the temperature is raised to 83.0Ā°C, what is the pressure of the gas remaining in the tank? 9.1e-21 Incorrect: Your answer is incorrect.

Respuesta :

Answer:

3.9 atm

Explanation:

Ideal gas law states:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The volume of the tank is constant, so we can say:

Vā‚ = Vā‚‚

Using ideal gas law to write in terms of P, n, R, and T:

nā‚ R Tā‚ / Pā‚ = nā‚‚ R Tā‚‚ / Pā‚‚

nā‚ Tā‚ / Pā‚ = nā‚‚ Tā‚‚ / Pā‚‚

Initially:

Pā‚ = 9.8 atm

Tā‚ = 29.0Ā°C = 302.15 K

nā‚ = n

Afterwards:

Pā‚‚ = P

Tā‚‚ = 83.0Ā°C = 356.15 K

nā‚‚ = n/3

Substituting:

n (302.15 K) / (9.8 atm) = (n/3) (356.15 K) / P

(302.15 K) / (9.8 atm) = (1/3) (356.15 K) / P

P = 3.85 atm

Rounding to 2 significant figures, P = 3.9 atm.