Question

A 3.20 L sample of neon at 4.57 atm is added to a 11.0 L cylinder that contains argon. If the pressure in the cylinder is 5.13 atm after the neon is added, what was the original pressure (in atm) of argon in the cylinder?

Answer 1

For neon gas using,

P_{​​​​​​1}​​​​​V_​1 = P_{​​​​​​2} V ₂

P ₁ = initial pressure of neon = 4.57 atm

V_{​​​​​​1} = initial volume = 3.2L

P_{​​​​​​2} = final pressure of neon

V_{​​​​​​2} = final volume= 11 L

Thus P_{​​​​​​2} = ( P_{​​​​​​1} V ₁)/V_{​​​​​​2}

P2. = (3.2L × 4.57atm)/11.0L

= 1.33 atm

Thus, the original pressure for argon gas will be = total pressure - P2

= (5.13- 1.33)atm

= 3.8 atm