Related rates with the ideal gas law

Starting from $PV = nRT$, differentiate both sides with respect to $t$, treating $P$, $V$, and $T$ as functions of time while $n$ and $R$ are constants. The left side requires the product rule: $\frac{d}{dt}(PV) = P\frac{dV}{dt} + V\frac{dP}{dt}$. This gives $P\frac{dV}{dt} + V\frac{dP}{dt} = nR\frac{dT}{dt}.$

Solve for $\frac{dT}{dt}$: $\frac{dT}{dt} = \frac{P\frac{dV}{dt} + V\frac{dP}{dt}}{nR}.$ Substituting $P = 10000$, $\frac{dV}{dt} = 0.3$, $V = 2$, $\frac{dP}{dt} = -50$, $n = 5$, and $R = 8.314$, the numerator is $10000(0.3) + 2(-50) = 3000 - 100 = 2900$, and the denominator is $5(8.314) = 41.57$. Thus $\frac{dT}{dt} \approx 69.76\ \text{K/s}$.

The temperature is increasing. In part a, dT/dt was computed to be approximately 69.76 kelvin per second, which is a positive value, meaning temperature rises at that instant even though pressure is falling, because volume is increasing quickly enough to dominate the combined effect.