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 $5.9 \mathrm {~L}$ What is the final pressure of the gas?
A)4.8 atm
B)4.3 atm
C)5.3 atm
D)5.8 atm
E)6.3 atm

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 know that the number of moles of gas is constant, so we can write:$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$where the subscripts 1 and 2 refer to the initial and final states of the gas, respectively.We can plug in the given values and solve for P_2:$$\frac{(5.6 	ext{ atm})(3.9 	ext{ L})}{330 	ext{ K}} = \frac{P_2(5.9 	ext{ L})}{430 	ext{ K}}$$$$P_2 = \frac{(5.6 	ext{ atm})(3.9 	ext{ L})(430 	ext{ K})}{(330 	ext{ K})(5.9 	ext{ L})} = 4.8 	ext{ atm}$$Therefore, the final pressure of the gas is 4.8 atm.

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