Answer:
[tex]3.59\times 10^{-21} J[/tex]
Explanation:
We are given that
Boltzmann's constant, [tex]k_B=1.38066\times 10^{-23} J/K[/tex]
Universal gas constant,R=8.31451 J/K
Number of moles,n=3
Volume ,V=6.1 L=[tex]6.1\times 10^{-3}m^3[/tex]
Pressure,P=7 atm=[tex]7\times 101325 Pa[/tex]
[tex]PV=nRT[/tex]
[tex]T=\frac{PV}{nR}=\frac{7\times 101325\times 6.1\times 10^{-3}}{3\times 8.31451}[/tex]
T=173.45 K
Average kinetic energy=[tex]\frac{3}{2}k_BT[/tex]
Average kinetic energy=[tex]\frac{3}{2}(1.38066\times 10^{-23}\times 173.45[/tex]
Average kinetic energy=[tex]3.59\times 10^{-21} J[/tex]
