Question

What is the temperature of a 3.5 mol of He gas contained in 34 L container with a pressure of 2.11 atm?

Answer 1

$PV= n RT$

$T= \frac{PV}{nR}$

P = 2.11 atm

n = 3.5

R = 0.0821 ( constant)

V = 34

$T=\frac{2.11×34}{3.5×0.082}=249.97 K$ or $-23.03°c$