An ideal gas is contained in a cylinder with a volume of \(5.0 \times 10^{2} \mathrm{mL}\) at a temperature of \(30 .^{\circ} \mathrm{C}\) and a pressure of \(710.\) torr. The gas is then compressed to a volume of \(25 \mathrm{mL}\) and the temperature is raised to \(820 .^{\circ} \mathrm{C}\). What is the new pressure of the gas?

In this problem, we are working with volumes in milliliters, pressure in torr, and temperature in Celsius. First, we need to convert them into the standard units to work with the ideal gas law: SI units for volume(liters), pressure(atmospheres), and temperature(Kelvin). Volume unit conversion: Initial volume, \(V_1 = 5.0 \times 10^2 \: mL = 0.500\: L\) Final volume, \(V_2 = 25\: mL = 0.025\: L\) Pressure unit conversion: Initial pressure, \(P_1 = 710.0\: torr\) We know that there are \(101325\: Pa\: (1\: atm)\) in \(760\: torr\). To convert the given pressure to atm, we divide by \(760\). \(P_1 = 710.0\: torr \times \frac{1\: atm}{760\: torr} = 0.9342\: atm\) Temperature unit conversion: Initial temperature, \(T_1 = 30.0\: ^\circ C\) We convert from Celsius to Kelvin by adding \(273.15\). \(T_1 = 30.0 + 273.15 = 303.15\: K\) Final temperature, \(T_2 = 820.0\: ^\circ C\) \(T_2 = 820.0 + 273.15 = 1093.15\: K\)

Now that we have converted the given values to appropriate units, we can use the combined gas law formula to find the new pressure: \(P_1V_1/T_1 = P_2V_2/T_2\) Rearrange equation to solve for the final pressure, \(P_2\): \(P_2 = \frac{P_1V_1T_2}{V_2T_1}\) Substitute the known values into the equation: \(P_2 = \frac{(0.9342\: atm)(0.500\: L)(1093.15\: K)}{(0.025\: L)(303.15\: K)}\) Calculate \(P_2\): \(P_2 = 76.674\: atm\) So, the new pressure of the gas after compression and heating is approximately \(76.7\: atm\).