A gas occupies 30.0dm 3 at S.T.P. What volume would it occupy at 91°C and...

Explanation

To answer this question, we need to apply the combined gas law, which relates the initial and final states of a gas sample when temperature, pressure, and volume change. The combined gas law formula is:

\[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]

Where \(P_1\) and \(P_2\) are the initial and final pressures, \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.

Given the initial conditions, we have \(V_1=30.0\,\text{dm}^3\) at STP (standard temperature and pressure). STP is defined as a temperature of \(0^\circ\text{C}\) (273.15 K) and a pressure of 760 mmHg. The final conditions are \(T_2 = 91^\circ\text{C}\) (364.15 K) and \(P_2 = 380\,\text{mmHg}\). We want to find \(V_2\).

First, let's rewrite the combined gas law formula to solve for \(V_2\):

\[V_2 = \frac{P_1V_1T_2}{P_2T_1}\]

Now plug in the given values and solve for \(V_2\):

\[V_2 = \frac{(760\,\text{mmHg})(30.0\,\text{dm}^3)(364.15\,\text{K})}{(380\,\text{mmHg})(273.15\,\text{K})}\]

Calculating the result, we get:

\[V_2 \approx 80.0\,\text{dm}^3\]

So, the correct answer isOption D: 80.0dm\(^3\).