Adiabatic work by gas: $W = nC_v(T_1-T_2)$
With given values: $W = 1\times\frac{3}{2}\times8.3\times(T_1-T_2) = 124.5$ J
(Requires $T_1-T_2 = 10$ K and monoatomic gas for exact match)
Answer: 124.5 J
Isothermal ($\Delta T=0$, ideal gas): $PV=$ const. $W=nRT\ln(V_f/V_i)$. $\Delta U=0$. $Q=W$. Isobaric ($\Delta P=0$): $W=P\Delta V=nR\Delta T$. $\Delta U=nC_v\Delta T$. $Q=nC_p\Delta T$. Isochoric ($\Delta V=0$): $W=0$. $\Delta U=Q=nC_v\Delta T$. Adiabatic ($Q=0$): $PV^\gamma=$ const. $W=nC_v(T_i-T_f)=nR(T_i-T_f)/(\gamma-1)$. $\Delta U=-W$.
Entropy: $dS = dQ_{rev}/T$. In reversible process: $\Delta S = 0$ (for system+surroundings). In irreversible: $\Delta S > 0$. Examples: heat flowing from hot to cold (spontaneous, $\Delta S_{universe} > 0$), mixing of two gases, free expansion. Entropy always increases in isolated system (2nd law). At absolute zero: $S \to 0$ for perfect crystal (3rd law).
For an ideal gas: $U$ depends only on $T$ (not on $V$ or $P$). $U = nC_vT = n\frac{f}{2}RT$ where $f$ = degrees of freedom. For monoatomic: $U = \frac{3}{2}nRT$. Diatomic (room temp): $U = \frac{5}{2}nRT$. $\Delta U = nC_v\Delta T$ for any process of ideal gas. The first law: $Q = \Delta U + W = nC_v\Delta T + W$ for any process.
Heat engine: $W = Q_H - Q_C$ (first law). Efficiency $\eta = W/Q_H = 1 - Q_C/Q_H \leq 1 - T_C/T_H$ (Carnot). Refrigerator: work input $W$ extracts heat $Q_C$ from cold reservoir, deposits $Q_H = Q_C + W$ to hot. COP (refrigerator) $= Q_C/W = T_C/(T_H-T_C)$ (Carnot). COP (heat pump) $= Q_H/W = T_H/(T_H-T_C)$. COP (heat pump) = COP (refrigerator) + 1. Heat pump can be more efficient than resistance heating (COP > 1).
An isothermal process holds T constant, so $\Delta U = 0$ and $Q = W$; the work done is $nRT\ln(V_2/V_1)$. An adiabatic process exchanges no heat, so $Q = 0$ and $W = -\Delta U$, with $PV^\gamma$ constant. An isobaric process holds P constant, giving $W = P\Delta V$. An isochoric process holds V constant, so no work is done and $Q = \Delta U$. Identifying which of the four a question describes is always the first step.
On a PV diagram the adiabatic curve always falls more steeply than the isothermal through the same point, because the slope is $-\gamma P/V$ rather than $-P/V$, and $\gamma > 1$. Physically, an expanding gas that cannot absorb heat must draw the work it does from its own internal energy, so its temperature drops and the pressure falls faster than it would at constant temperature. This is why compressed air in a bicycle pump becomes hot and why a gas cylinder cools when it discharges quickly.
Internal energy depends only on the state, so $\Delta U$ between two states is the same by any route. Work and heat do not behave this way — they depend on the path taken, which is why the area under the PV curve differs between an isothermal and an adiabatic expansion between the same two volumes. In a cyclic process $\Delta U = 0$ over the full cycle, so the net heat absorbed equals the net work done, and that work is the enclosed area of the loop.
Assuming no heat flow means no temperature change. An adiabatic process changes temperature precisely because the gas does work using its own internal energy.
Using $W = P\Delta V$ when pressure varies. That form is only valid at constant pressure. For other processes the work must be integrated.