Rational Expressions Word problem on combined variation The volume $V$ of a fixed amount of a gas varies directly as the temperature $T$ and inversely as the pressure $P$. Suppose that $V=70 \mathrm{~cm}^{3}$ when $T=420$ kelvin and $P=18 \frac{\mathrm{kg}}{\mathrm{cm}^{2}}$. Find the pressure when $T=140$ kelvin and $V=60 \mathrm{~cm}^{3}$. \[ \square \frac{\mathrm{kg}}{\mathrm{cm}^{2}} \] $\times$ 5
Solution
Step 1 ：The problem states that the volume of a gas varies directly with the temperature and inversely with the pressure. This can be represented by the equation: \(V = \frac{kT}{P}\) where k is the constant of variation.
Step 2 ：We can find the value of k using the given conditions \(V = 70 cm³\), \(T = 420 K\), and \(P = 18 kg/cm²\). Substituting these values into the equation gives: \(70 = \frac{k(420)}{18}\).
Step 3 ：Solving for k gives: \(k = \frac{70 * 18}{420} = 3\).
Step 4 ：Now we can use this value of k to find the pressure when \(T = 140 K\) and \(V = 60 cm³\). Substituting these values into the equation gives: \(60 = \frac{3(140)}{P}\).
Step 5 ：Solving for P gives: \(P = \frac{3 * 140}{60} = 7 kg/cm²\).
Step 6 ：So, the pressure when \(T = 140 K\) and \(V = 60 cm³\) is \(\boxed{7 kg/cm²}\).
