Question

Calculate total energy of one mole of an ideal monatomic gas at 27°c

Answer 1

Since the process is adiabatic, so, q=0q=0

As the gas expands against the constant external pressure

W=−P(ΔV)=−1.5×(V2–V1)=−1.5(16−4)W=−P(ΔV)=−1.5×(V2–V1)=−1.5(16−4)

=−18atmdm3=−18×101.325J=−1823.85/j=−18atmdm3=−18×101.325J=−1823.85/j

So, ΔE=q+W=0+(−1823.85)

=−1823.85J