Question

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in a 9.0 L flask. The total pressure in the flask was _____ atm. Assume the initial pressure in the flask was 0.00 atm.

Answer 1

Answer:

3.6667

Explanation:

For helium gas:

Using Boyle's law  

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${5.6}\times {3.0}={P_2}\times {9.0} atm$

${P_2}=\frac {{5.6}\times {3.0}}{9.0} atm$

${P_1}=1.8667\ atm$

The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm

For Neon gas:

Using Boyle's law  

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${3.6}\times {4.5}={P_2}\times {9.0} atm$

${P_2}=\frac {{3.6}\times {4.5}}{9.0} atm$

${P_1}=1.8\ atm$

The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm

Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.