Asample of neon gas at 1.20 atm compresses from 0.250 l to 0.125 l. if the temperature remains constant, what is the final pressure in atm? a. 0.600 atm b. 1.000 atm c. 1.20 atm d. 2.40 atm e. none of the above



2.4 atm


At constant temperature;

PV = constant  

P1 = 1.2 atm

V1 = 0.25 L

P2 = ?

V2 = 0.125 L


P1*V1 = P2*V2


P2 = P1*V1/V2

     = (1.2 atm)(0.25 L)/(0.125 L)

    = 2.4 atm

D. 2.40 atm


At constant temperature and number of moles, using Boyle's law  

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

Given ,  

V₁ = 0.250 L

V₂ = 0.125 L

P₁ = 1.20 atm

P₂ = ? atm

Using above equation as:

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

{1.20\ atm}\times {0.250\ L}={P_2}\times {0.125\ L}

{P_2=\frac{{1.20}\times {0.250}}{0.125}\ atm

{P_2}=2.40 atm

- D. 2.40 atm

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