Question

A 3.20-mol sample of gas occupies a volume of 350. mL at 300.0 K. Determine the

Answer 1

Answer:

$P \approx 225atm$

Explanation:

From the question we are told that:

Moles of sample n=3.20_mol

Volume V=350mL

Temperature T=300k

Generally the equation for ideal gas is mathematically given by

 $PV=nRT$

 $P=\frac{nRT}{V}$

 $P=\frac{3.20*0.08206*300}{350*10^{-3}}$

 $P=225.079atm$

 $P \approx 225atm$