One mole of an ideal gas with a volume of 4.00 liters and a temperature of 0$^{\circ}$C is mixed with one mole of an ideal gas with a volume of 2.00 liters and a temperature of 100$^{\circ}$C.The volume of the mixture is 6 liters.What is the N/V of the mixture? A) 1.75 $\times$ 10²⁶ m^-3 B) 2.01 $\times$ 10²⁶ m^-3 C) 2.66 $\times$ 10²⁶ m^-3 D) 3.21 $\times$ 10²⁶ m^-3 E) 3.56 $\times$ 10²⁶ m^-3

Question

Quizplus · Accepted Answer

First, we need to convert the temperatures to Kelvin:
T1 = (0°C + 273.15) K = 273.15 K
T2 = (100°C + 273.15) K = 373.15 K

Then, we can use the combined gas law to find the N/V of the mixture:
(P1 V1 / N1 T1) = (P2 V2 / N2 T2)

We know that P1 = P2 (since they are mixed together), so we can simplify the equation to:
V1/N1T1 = V2/N2T2

Substituting the given values:
(4 L / 1 mol) / (1 mol / 273.15 K) = (2 L / 1 mol) / (1 mol / 373.15 K)

Simplifying:
N = V1T2 / (V2T1) = (4 L)(373.15 K) / (2 L)(273.15 K) = 2.01 x 10^-3 mol/L

Therefore, the answer is B: 2.01 x 10^-3 mol/L.

One Mole of an Ideal Gas with a Volume of 4.00