Question

You are given a 3.00 mol sample of krypton with a volume of 7.00 L. The temperature is held at 300. K. What is the pressure of the krypton sample? Show your work to receive full credit.
i need help ASAP

Answer 1

1) list your givens

P = ?

V = 7.00 L

n = 3.00 mol

R = 8.314 $\frac{kPa\ L}{mol\ k}$

T= 300.15 K

2) rearrange your formula

$PV=nRT\\\\P=\frac{nRT}{V}$

3) solve

$P=\frac{(3.00\ mol)(8.314\ \frac{kPa\ L}{mol\ k})(300.15\ K)}{(7.00\ L)}\\\\P=1069.4773299$

$P=1.07$ × $10^{3}\ kPa$