Two moles of a monoatomic ideal gas at 300 K and 10 atm expand adiabatically against a constant external pressure of 1 atm until the pressure of the…

The process is irreversible, so the reversible adiabatic formula cannot be used for the final temperature. The first law gives $nC_v(T_2-T_1)=-P_{ext}(V_2-V_1)$ with $V=nRT/P$. With $C_v=\tfrac32R$: $\tfrac32(T_2-300)=-(T_2-0.1\times300)$, i.e. $2.5T_2=450+30$ and $T_2=192$ K. The entropy is a state function, so we evaluate it along a reversible path: $\Delta S=nC_p\ln(T_2/T_1)+nR\ln(P_1/P_2)$. This is $2(2.5)(8.314)\ln(192/300)+2(8.314)\ln10=41.57(-0.4463)+38.29=-18.55+38.29=19.74\ \mathrm{J\,K^{-1}}$. It is positive, as it must be for an irreversible adiabatic process, since the entropy of the surroundings does not change and the total entropy must increase. Note that the reversible-adiabatic result would be $\Delta S=0$ with a lower final temperature of 119 K.