Gas is confined in a tank at a pressure of 9.8 atm and a temperature of 29.0°C. If two-thirds of the gas is withdrawn and the temperature is raised to 83.0°C, what is the pressure of the gas remaining in the tank? 9.1e-21 Incorrect: Your answer is incorrect.

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:

V₁ = V₂

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

n₁ R T₁ / P₁ = n₂ R T₂ / P₂

n₁ T₁ / P₁ = n₂ T₂ / P₂

Initially:

P₁ = 9.8 atm

T₁ = 29.0°C = 302.15 K

n₁ = n

Afterwards:

P₂ = P

T₂ = 83.0°C = 356.15 K

n₂ = 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.