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if a sample of gas is intially at 1.8 atm,22.0 l, and 26.4 c, what will be the volume if the pressure is reduced by 0.8 atm and the temperature is lowered to 20.3 c?

Respuesta :

Kalahira
pv=nrT Initial state (1.8atm)(22.0 l)=n(0.082057)(26.4+273.15); r=.082057, and converting C to K Solving for n = (1.8)(22)/(.082057*(26.4+273.15) moles n = 1.611 moles in initial state Now we solve for new volume pv=nrT (.8atm)v=(1.611)(.082057)(20.3+273.15) v=(1.611)(.082057)(20.3+273.15)/.8 v=48.49 l
transitionstate

Answer:

              48.49 L

Solution:

Data Given:

                 P₁  =  1.8 atm

                 V₁  =  22.0 L

                 T₁  =  26.4 °C  =  299.55 K

                 P₂  =  0.8 atm

                 V₂  =  ??

                 T₂  =  20.3 °C  =  293.45 K

Formula Used:

Let's assume that the gas is acting as an Ideal gas, then according to Ideal Gas Equation,

                  P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving Equation for V₂,

                  V₂  =  P₁ V₁ T₂ / T₁ P₂

Putting Values,

                  V₂  =  (1.8 atm × 22.0 L × 293.45 K) ÷ (299.55 K × 0.8 atm)

                  V₂  =  48.49 L

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