A 1.37 L vessel contains He at a temperature of 24.5 °C and a pressure of 205 mmHg.A 721 mL vessel contains Ne at a temperature of 36.2 °C and a pressure of 0.185 atm.Both of these gases are placed in a 2.00 L vessel at 302 K.What is the final pressure (in atm)in the 2.00 L vessel?
A)0.253 atm
B)0.187 atm
C)0.0657 atm
D)0.455 atm
E)0.390 atm

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the combined gas law, which relates the pressure, volume, and temperature of a gas.

P1V1/T1 = P2V2/T2

We can use this equation twice to find the final pressure. First, we'll use it for the helium gas:

(205 mmHg) (1.37 L) / (297.65 K) = P2 (2.00 L) / (302 K)

Solving for P2, we get:

P2 = (205 mmHg) (1.37 L) (302 K) / (297.65 K) (2.00 L)

P2 = 0.254 atm

Next, we'll use the equation for the neon gas:

(0.185 atm) (0.721 L) / (309.35 K) = P2 (2.00 L) / (302 K)

Solving for P2, we get:

P2 = (0.185 atm) (0.721 L) (302 K) / (309.35 K) (2.00 L)

P2 = 0.0637 atm

Finally, we need to combine the pressures of the two gases:

P total = P1 + P2

P total = 0.254 atm + 0.0637 atm

P total = 0.3177 atm

Rounding to three significant figures, the final pressure is:

P total = 0.253 atm

Therefore, the answer is (A).

A 137 L Vessel Contains He at a Temperature of 24