The temperature of 0.52 mol of gas at a pressure of 1.3 atm and a volume of 11.8L is ____K.?

Question

The temperature of 0.52 mol of gas at a pressure of 1.3 atm and a volume of 11.8L is ____K.

What is the answer in scentific notation with sig. figs.?

Answer 1

(0.52 mol)(22.4 L/mol) = 11.648 L @ STP

(1 atm)(11.648 L)/(293.15° K) = (1.3 atm)(11.6 L)/T

T = ((1.3 atm)/(1 atm))((11.6 L)/(11.648 L))(293.15° K)

T ≈ 379.52° K ≈ 3.8 x 10^2° K