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A certain gas occupies a volume of 4.3Lat a pressure of 1.2atmand a temperature of 310K. It is compressed adiabatically to a volume of 0.76L

Determine

  1. The final pressure
  2. The final temperature, assuming the gas to be an ideal gas for which=1.4

Short Answer

Expert verified
  1. The final pressure is 14 atm.
  2. The final temperature is 6.2102K.

Step by step solution

01

Given data

  • Initial volume isVi=4.3L
  • Final volume is Vf=0.76L
  • Initial pressure is pi=1.2atm
  • Temperature is Ti= 310K
  • =1.4
02

Understanding the concept

In case of adiabatic process,

pV=Constant

Here p is the pressure, V is the volume and is the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

localid="1662398424168" =CPCV

Here CP is specific heat capacity at constant pressure and CVis specific heat capacity at constant volume.

03

(a) Calculate the final pressure

For adiabatic process, we can write

piV=pfVf

role="math" localid="1662442278944" piViVf=pf

Substitute 1.2 atm for pi , 4.3L for Vi , 0.76L for Vfinto the above equation,

pf=1.24.30.761.4=13.5799atm=14atm

Therefore the final pressure is 14atm .

04

(b) Calculate the final temperature

We can write two equations according to the gas law as

piVi=nRTi...........(i)
pfVf=nRTf..... (ii)

Dividing second equation by the first, we get

pfVfpiVi=nRTfnRTi

Substitute 13.5799 atm for pf , 0.76L for Vf ,1.2atm for pi, 4.3L for Vi and 310K for TI into the above equation,

Tf=13.57990.761.24.3310=6.2102K

Therefore the final temperature is 6.2102K.

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