Question

The stopcock connecting a 1.22 L bulb containing helium gas at a pressure of 3.41 atm, and a 6.67 L bulb containing Xenon gas at a pressure in the system is _______atm.

Use the References to access important values if needed for this question. The stopcock connecting a 1.22 L bulb containing helium gas at a pressure of 3.41 atm, and a 6.67 L bulb containing xenon gas at a pressure of 3.90 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is atm

Answer 1

He = 10 22 Lith PHe = 3.41 codony ## / Vxe = 6:676) Pxe = 3.gute put xe He The = Pue x Vie nxe = Pxex Uxe RT The = 3.41*1022 RT nxe = 3.9x6.67 RT Now stopcock is removed RT Prece . Prewe Treel He xe Vtoteel = 1-22 + 6.67 = 7.89 Lt Prew = ? Bulbs are closed so toteed moles will be constens total = nHe thxe thero Vitotul = 3.41x1-22 + 3.9 x6:67 RT RT RT Precy (789) 4.1602 + 26.013 Prem = 3.82 atm) Ang

Answer 2

SOLUTION :

P*V of helium =  = 3.41 * 1.22 = 4.1602 L-atm

P*V of xenon = 3.90 * 6.67 = 26.0130  L-atm

When these two bulbs are connected through a cock,

P*V of the system = Total of the P*V of the two bulbs = 4.1602 + 26.0130 = 30.1732 L-atm

=> P of the system * V of the system = 30.1732

=> P of the system = 30.1732 / V of the system

=> P of the system = 30.1732 / (1.22 + 6.67)

=> P of the system = 3.8242 atm 

Final pressure in the system is 3.8242 atm (ANSWER)