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Consider a gas at 1.00 atm in a 5.00-L container at 20.0 °C. What pressure does the gas exert when transferred to a volume of 2.00 L at 43 °C?

User CBaker
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1 Answer

3 votes

Answer:

The pressure is 2.696 atm

Step-by-step explanation:

Boyle's law relates pressure and volume inversely proportionally and is expressed mathematically as:

P * V = k

Charles's law consists of the relationship between the volume and temperature of a certain amount of ideal gas, which is maintained at a constant pressure, by means of a proportionality constant that is applied directly and is expressed mathematically as follows:


(V)/(T) =k

Gay-Lussac's law indicates that the gas pressure is directly proportional to its temperature and is expressed mathematically as:


(P)/(T) =k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:


(P*V)/(T) =k

When you want to study two different states, an initial and a final one of a gas, you have the expression:


(P1*V1)/(T1) =(P2*V2)/(T2)

In this case, you know:

  • P1=1 atm
  • V1= 5 L
  • T1= 20 °C=293° K (Being 0°C=273°K)
  • P2=?
  • V2= 2 L
  • T2= 43 °C= 316°K

Replacing:


(1 atm*5 L)/(293K) =(P2*2L)/(316K)

Solving:


P2=(1 atm*5 L)/(293K) *(316 K)/(2L)

P2= 2.696 atm

The pressure is 2.696 atm

User Antonio Glavocevic
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