Question

A constant-volume tank contains a mixture of \(120 \mathrm{g}\) of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(600 \mathrm{g}\) of \(\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(200 \mathrm{kPa} .\) The contents of the tank are now ignited, and the methane gas burns completely. If the final temperature is \(1200 \mathrm{K},\) determine \((a)\) the final pressure in the tank and ( \(b\) ) the heat transfer during this process.

Short answer

In this exercise, we calculated the final pressure and heat transfer for a combustion reaction in a constant-volume tank containing methane and oxygen. We first determined the mole numbers of each constituent and balanced the reaction. Then, we applied the ideal gas law to find the final pressure, which came out to be approximately 2113.77 kPa. Finally, we used the energy balance equation to calculate the heat transferred during the process, which was approximately 4.69 * 10^6 J.

Step-by-step solution

List the given data and determine mole numbers

Given data: Mass of methane (CH4): 120 g Mass of oxygen (O2): 600 g Initial Temperature (T1): 25 °C = 298.15 K Initial Pressure (P1): 200 kPa Final Temperature (T2): 1200 K Using the molar masses, we will determine the moles of methane and oxygen: Molar mass of CH4: 16 g/mol Molar mass of O2: 32 g/mol Moles of CH4 (n1) = (120 g) / (16 g/mol) = 7.5 mol Moles of O2 (n2) = (600 g) / (32 g/mol) = 18.75 mol

Perform stoichiometric balancing of the combustion reaction

The equation for the complete combustion of methane with oxygen is given below: CH4 + 2 O2 -> CO2 + 2 H2O From the stoichiometric ratio, we can determine the amount of oxygen consumed and the amount of products formed: 7.5 mol of CH4 reacts with 15 mol of O2 (Since 2 moles of O2 are required for 1 mole of CH4) This reaction produces 7.5 mol of CO2 and 15 mol of H2O. Excess O2 present = initial moles of O2 - consumed moles of O2 = 18.75 - 15 = 3.75 mol

Apply the ideal gas law to determine the final pressure

Since the volume of the tank is constant, we can use the ideal gas law in the form: n T2/T1 = P2/P1 (rearranging: P2 = n T2/T1 * P1) The total moles of the gas mixture at the final state (n) = Moles of CO2 + Moles of H2O + Excess O2 n = 7.5 + 15 + 3.75 = 26.25 mol Now, we can apply the ideal gas law equation: P2 = (26.25 mol) * (1200 K) / (298.15 K) * (200 kPa) P2 ≈ 2113.77 kPa Therefore, the final pressure in the tank is approximately 2113.77 kPa.

Determine the heat transfer during the process

To determine the heat transfer, we apply the energy balance equation. Q - W = ΔU Since it is a constant volume process, there is no work done (W=0). Then, Q = ΔU Internal energy change ΔU is obtained as: ΔU = n * cv,m * ΔT where cv,m is the molar specific heat capacity at constant volume, and ΔT is the temperature change. For the combustion reaction, we have: ΔU = (ΔU_CH4 + ΔU_O2 + ΔU_CO2 + ΔU_H2O) Assuming ideal gas behavior, we have for each gas component: ΔU_gas = nf * cv,gas * (T2 - T1) where nf is the final number of moles of each gas component. To account for all the components, we sum the energy changes for CH4 (consumed), O2 (initial excess and consumed summed), CO2 (produced), and H2O (produced). Using molar specific heat capacities, cv,m (J/mol K), for each component: cv,m(CH4) = 35.69 cv,m(O2) = 29.41 cv,m(CO2) = 33.59 cv,m(H2O) = 33.15 Now, we calculate the total internal energy change: ΔU = [(-7.5 mol) * 35.69 + (-15 mol) * 29.41 + (7.5 mol) * 33.59 + (15 mol) * 33.15] * (1200 K - 298.15 K) ΔU ≈ 4.69 * 10^6 J Since Q = ΔU, the heat transferred during the process is approximately 4.69 * 10^6 J.