Reversible and Irreversible adiabatic expansion
The change in internal energy $U$ is
$$\Delta U=Q+W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system.
Since the process is adiabatic, no heat is transferred into or out of the system, i.e. $Q=0$ and thus
$$\Delta U=W$$

The reversible expansion is performed continuously at equilibrium by means of infinitesimal changes in pressure $p$.
The corresponding infinitesimal pressure-volume work $\delta W$ is
$$\delta W=-p\,\mathrm dV$$

For an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$. Since $n$ is constant in a closed system, the change in internal energy $U$ is
$$\mathrm dU=C_V\,\mathrm dT$$
where $C_V$ is heat capacity at constant volume.

Equating $\mathrm dU$ and $\delta W$ yields
$$C_V\,\mathrm dT=-p\,\mathrm dV$$
and since $pV=nRT$ for an ideal gas
$$\begin{align}
C_V\,\mathrm dT&amp;=-\frac{nRT}V\,\mathrm dV\\
\frac{C_V}T\,\mathrm dT&amp;=-\frac{nR}V\,\mathrm dV
\end{align}$$

Integration from the initial state ($T_1,V_1$) to the final state ($T_2,V_2$) yields
$$\begin{align}
\int\limits_{T_1}^{T_2}\frac{C_V}T\,\mathrm dT&amp;=\int\limits_{V_1}^{V_2}-\frac{nR}V\,\mathrm dV\\
C_V\int\limits_{T_1}^{T_2}\frac1T\,\mathrm dT&amp;=-nR\int\limits_{V_1}^{V_2}\frac1V\,\mathrm dV\\
C_V\ln\left(\frac{T_2}{T_1}\right)&amp;=-nR\ln\left(\frac{V_2}{V_1}\right)\\
\ln\left(\frac{T_2}{T_1}\right)&amp;=-\frac{nR}{C_V}\ln\left(\frac{V_2}{V_1}\right)\\
\frac{T_2}{T_1}&amp;=\left(\frac{V_2}{V_1}\right)^{-nR/C_V}\\
&amp;=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}
\end{align}$$
where $C_{\mathrm m,V}$ is molar heat capacity at constant volume.

Since $C_{\mathrm m,p}-C_{\mathrm m,V}=R$, the exponent can be rewritten
$$\begin{align}
\frac{T_2}{T_1}&amp;=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}\\
&amp;=\left(\frac{V_1}{V_2}\right)^{C_{\mathrm m,p}/C_{\mathrm m,V}-1}\\
&amp;=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
\end{align}$$
where $\gamma$ is the ratio of the heat capacities $\gamma=C_p/C_V=C_{\mathrm m,p}/C_{\mathrm m,V}=c_p/c_V$. For a monoatomic ideal gas, $C_{\mathrm m,p}=\tfrac52R$ and $C_{\mathrm m,V}=\tfrac32R$, and thus $\gamma=\tfrac53$.

Note that, in the given question, the volumes $V_1$ and $V_2$ are unknown. However, the values for the pressures $p_1$ and $p_2$ are given. Since $pV=nRT$ for an ideal gas,
$$\begin{align}
\frac{T_2}{T_1}&amp;=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
T_2\cdot V_2^{\gamma-1}&amp;=T_1\cdot V_1^{\gamma-1}\\
\frac{p_2V_2}{nR}\cdot V_2^{\gamma-1}&amp;=\frac{p_1V_1}{nR}\cdot V_1^{\gamma-1}\\
p_2\cdot V_2\cdot V_2^{\gamma-1}&amp;=p_1\cdot V_1\cdot V_1^{\gamma-1}\\
p_2\cdot V_2^\gamma&amp;=p_1\cdot V_1^\gamma\\
\frac{p_2}{p_1}&amp;=\left(\frac{V_1}{V_2}\right)^\gamma
\end{align}$$

In contrast to the reversible expansion, an irreversible expansion is not performed continuously at equilibrium by means of infinitesimal changes in pressure. In the limiting case, the value of the pressure $p$ abruptly changes from $p_1$ and $p_2$. The following expansion proceeds in one step against a constant pressure $p_2$. Hence,
$$\delta W=-p_2\,\mathrm dV$$
and thus
$$\begin{align}
C_V\,\mathrm dT&amp;=-p_2\,\mathrm dV\\
\int\limits_{T_1}^{T_2}C_V\,\mathrm dT&amp;=\int\limits_{V_1}^{V_2}-p_2\,\mathrm dV\\
C_V\left(T_2-T_1\right)&amp;=-p_2\left(V_2-V_1\right)
\end{align}$$
Since $pV=nRT$ for an ideal gas,
$$\begin{align}
C_V\left(T_2-T_1\right)&amp;=-p_2\left(V_2-V_1\right)\\
&amp;=-p_2\left(\frac{nRT_2}{p_2}-\frac{nRT_1}{p_1}\right)\\
&amp;=-nRT_2+nRT_1\frac{p_2}{p_1}\\
C_VT_2+nRT_2&amp;=C_VT_1+nRT_1\frac{p_2}{p_1}\\
T_2\left(C_V+nR\right)&amp;=T_1\left(C_V+nR\frac{p_2}{p_1}\right)\\
T_2\left(C_{\mathrm m,V}+R\right)&amp;=T_1\left(C_{\mathrm m,V}+R\frac{p_2}{p_1}\right)\\
\end{align}$$