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 use the gas laws, specifically the combined gas law, which relates the initial and final states of a gas in terms of its pressure, volume, and temperature. The combined gas law is given by:

\( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)

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.

First, we need to convert the given values to the appropriate units. Since the initial conditions are at S.T.P (Standard Temperature and Pressure), we can use the standard values for pressure and temperature:

Initial Pressure (\(P_1\)) = 760 mmHg (Standard pressure)

Initial Volume (\(V_1\)) = 30.0 dm\(^3\)

Initial Temperature (\(T_1\)) = 273 K (Standard temperature)

Now let's convert the final conditions to the appropriate units:

Final Pressure (\(P_2\)) = 380 mmHg

Final Temperature (\(T_2\)) = 91