= Solution
Let $(T_1,V_1)$ be the final equilibrium state. The suddenly added weight supplies the constant external pressure $p_1$ during the displacement. Since the process is adiabatic, the first law gives
$$
\frac52Nk_B(T_1-T_0)=p_1(V_0-V_1).
$$
The initial and final ideal-gas equations are
$$
p_0V_0=Nk_BT_0,
\qquad
p_1V_1=Nk_BT_1.
$$
Substitution yields
$$
\frac52(T_1-T_0)=\frac{p_1}{p_0}T_0-T_1,
$$
and hence
$$
T_1=T_0\frac{5+2p_1/p_0}{7}.
$$
Therefore
$$
\boxed{\Delta T=T_1-T_0
=\frac{2T_0}{7}\left(\frac{p_1}{p_0}-1\right).}
$$
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