A 3.9-L volume of ideal neon gas (monatomic)is at a pressure of 5.6 aym and a temperature of $330 \mathrm {~K} .$ The atomic mass of neon is $20.2 \mathrm {~g} / \mathrm { mol }$ The temperature of the gas is now increased to 430 K and the volume is increased to $5.9 \text { L. }$ What is the final pressure of the gas?

Question

Quizplus · Accepted Answer

Using the ideal gas law, we can write:$$PV = nRT$$where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.We can rearrange this equation to solve for P:$$P = \frac{nRT}{V}$$We can now plug in the given values:$$n = \frac{3.9 	ext{ L}}{22.4 	ext{ L/mol}} = 0.174 	ext{ mol}$$$$R = 0.0821 	ext{ L atm/(mol K)}$$$$T_1 = 330 	ext{ K}$$$$V_1 = 3.9 	ext{ L}$$$$T_2 = 430 	ext{ K}$$$$V_2 = 5.9 	ext{ L}$$We can now solve for P:$$P = \frac{(0.174 	ext{ mol})(0.0821 	ext{ L atm/(mol K)})(430 	ext{ K})}{5.9 	ext{ L}} = 4.8 	ext{ atm}$$Therefore, the final pressure of the gas is 4.8 atm.

A 39-L Volume of Ideal Neon Gas (Monatomic)is at a Pressure