Answer: The work done on the gas mixture is [tex]+6.99\times 10^3Joules[/tex]
Explanation:
According to first law of thermodynamics:
[tex]\Delta E=q+w[/tex]
[tex]\Delta E[/tex]=Change in internal energy
q = heat absorbed or released
w = work done on or by the system
w = work done on the system =[tex]-P\Delta V[/tex] {Work done on the system is positive when the final volume is lesser than initial volume }
w =[tex]-69.0atm\times (98.0-99.0)L=+69.0Latm[/tex]
w =[tex]+69.0Latm\times 101.3=+6.99\times 10^3Joules[/tex] {1Latm=101.3J}
The work done on the gas mixture is [tex]+6.99\times 10^3Joules[/tex]
