Question

A cylinder contains 3.2 L of oxygen at 350 K and 2.5 atm . The gas is heated, causing a piston in the cylinder to move outward. The heating causes the temperature to rise to 600 K and the volume of the cylinder to increase to 9.1 L .

Answer 1

`P_(2)=1.8atm`

Step-by-step explanation:

The question molded an ideal gas law, which state in part

"The Volume(V) of a gas is directly proportional to the absolute temperature(T) and inversely proportional to the pressure(P)"

Mathematically, this can be expressed as

`V\alpha (T)/(P)\\V=K (T)/(P)\\`

where K is a constant. varying the value of k, we arrive at

`(P_(1) *V_(1))/(T_(1))=(P_(2) *V_(2))/(T_(2))=...=(P_(n) *V_(n))/(T_(n))\\`

using the first two expression of the above equation, where

`P_(1)= 2.5atm, V_(1)=3.2L, T_(1)= 350K \\P_(1)= X,V_(2)= 9.1atm,T_(2)= 600K\\`

Note, we have to convert the volume to
`m^(3) \\`

since
`1L=10^(-3)m \\`

Now
`V_(1)=3.2*10^(-3)=\\V_(2)=9.1*10^(-3)=0.0091m^(3) \\`

also we convert the pressure from atm to pascal(pa)
`1atm=1.013*10^(5) pa\\`

`P_(1)=2.5*1.013*10^(5)=2.53*10^(5)\\`

`P_(2)=(2.53*10^(5)pa*0.0032m^(3)*600)/(0.0091*350)\\`

`P_(2)=(4.858*10^(5))/(2.73) =1.78*10^(5)\\`

convert back to atm

`P_(2)=(1.788*10^(5))/(1.013*10^(5))=1.76atm`

Hence the value of he final pressure is

1.8atm