An ideal gas at 25.8°C and a pressure 1.20 x 10^5 Pa is in a container having a volume of 1.00 L. (a) Determine the number of moles of gas in the container. (b) The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature.

a) 0.0483 mol

b) 232 °C

Explanation:

Ideal gas law:

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.

a) Given:

P = 1.20*10⁵ Pa

V = 1.00 L = 1.00*10⁻³ m³

T = 25.8 °C = 298.95 K

PV = nRT

(1.20*10⁵ Pa) (1.00*10⁻³ m³) = n (8.314 m³ Pa / mol / K) (298.95 K)

n = 0.0483 mol

b) Given:

P = 1.013*10⁵ Pa

V = 2.00 L = 2.00*10⁻³ m³

n = 0.0483 mol

PV = nRT

(1.013*10⁵ Pa) (2.00*10⁻³ m³) = (0.0483 mol) (8.314 m³ Pa / mol / K) T

T = 505.73 K

T = 232 °C