Using Boyle's law to find the rate of volume change

Expert intro: Boyle's law links pressure and volume through the constant product $PV=C$, so a related-rates derivative is the right tool here.

Detailed walkthrough

Key idea: Boyle's law and related rates

Boyle's law says that for a fixed amount of gas at constant temperature, the product of pressure and volume stays constant: $PV=C$. Because $P$ and $V$ both change with time, we differentiate with respect to time instead of trying to solve the equation as a static algebraic identity.

At the instant given, the gas has $V=450\text{ cm}^3$, $P=150\text{ kPa}$, and the pressure is increasing at $15\text{ kPa/min}$. The quantity we want is $\frac{dV}{dt}$, the rate at which the volume is changing.

Differentiate the equation with respect to time

Starting from $PV=C$, differentiate both sides with respect to $t$:

$\frac{d}{dt}(PV)=\frac{d}{dt}(C)=0$

Using the product rule:

$P\frac{dV}{dt}+V\frac{dP}{dt}=0$

Now substitute the known values $P=150$, $V=450$, and $\frac{dP}{dt}=15$:

$150\frac{dV}{dt}+450(15)=0$

$150\frac{dV}{dt}+6750=0$

$150\frac{dV}{dt}=-6750$

$\frac{dV}{dt}=-45\text{ cm}^3/\text{min}$

Interpret the sign and units

The negative sign means the volume is decreasing, which matches the physical situation of compression. So the volume is decreasing at a rate of $45\text{ cm}^3/\text{min}$.

The final answer should be stated clearly as a rate of decrease, not just a negative number, because that is how the wording of the question is framed.

Common checking point

Because pressure is given in kPa and volume in cm^3, the rate from the differentiated equation inherits units of cm^3/min when pressure is changing in kPa/min. There is no need to convert units unless the question asks for a specific system. The essential relation is the derivative of the product $PV$.

So, the volume is decreasing at $45\text{ cm}^3/\text{min}$.

💡 Pitfall guide

A frequent mistake in $PV=C$ questions is to plug the numbers straight into $P+V=C$ or to differentiate only one variable and forget the other one changes too. Here, $P$ is increasing, so both $P$ and $V$ depend on time, and the product rule is unavoidable. Another common error is losing the minus sign: if the pressure goes up while the product stays constant, the volume must go down. Also watch the units. If pressure is in kPa and volume is in cm^3, then $\frac{dV}{dt}$ should be reported in cm^3 per minute, not just as a bare number. Finally, do not substitute the rate values before differentiating; the derivative equation must be written first, otherwise the algebra becomes incorrect and the physical meaning is lost.

🔄 Real-world variant

If the same Boyle's law setup had $V=600\text{ cm}^3$, $P=120\text{ kPa}$, and $\frac{dP}{dt}=10\text{ kPa/min}$, the equation would still be $PV=C$. Differentiating gives $P\frac{dV}{dt}+V\frac{dP}{dt}=0$, and substituting the new values gives $120\frac{dV}{dt}+600(10)=0$. That leads to $\frac{dV}{dt}=-50\text{ cm}^3/\text{min}$. If the pressure were decreasing instead, say $\frac{dP}{dt}=-10\text{ kPa/min}$, then the same method would produce a positive $\frac{dV}{dt}$, meaning the gas is expanding. This shows how the sign of the rate depends on whether the gas is being compressed or allowed to expand.

🔍 Related terms

related rates in gases, Boyle's law derivative, product rule physics