Answer:
Explanation:
From the question we are told that:
Energy [tex]Q=7950kcal=3.3*10^7[/tex]
Initial Volume [tex]V_1=12.8 m^2[/tex]
Final Volume [tex]V_2=19.4 m^2[/tex]
a)
Generally the equation for Work done is mathematically given by
[tex]W=P \triangle V[/tex]
Where
[tex]P= Pressure at 1atm[/tex]
Therefore
[tex]W=(1.01*10^5)(19.4-12.8)[/tex]
[tex]W=6.67*10^5J[/tex]
a)
Generally the equation for Change in internal energy of the gas is mathematically given by
[tex]\triangle U=Q-W[/tex]
[tex]\triangle U=3.3*10^7-6.67*10^5J[/tex]
[tex]\triangle U=3.2*10^7J[/tex]