1.000 mol of urea is slowly combusted in a container with a mobile piston. As a result of the combustion, the system expands quasistatically against an external pressure of 2.0 atm as it lifts a light weight. The temperature of the system is maintained at 298.15 K
The correct answer and explanation is:
To analyze this process, we need to apply the First Law of Thermodynamics and understand the thermochemical reaction involved in the combustion of urea (CO(NH₂)₂).
Combustion Reaction of Urea:
The balanced combustion reaction for 1 mole of urea is approximately: CO(NH2)2(s)+32O2(g)→CO2(g)+N2(g)+2H2O(l)\text{CO(NH}_2\text{)}_2 (s) + \frac{3}{2} O_2 (g) \rightarrow CO_2 (g) + N_2 (g) + 2 H_2O (l)
At 298.15 K, water forms in the liquid state.
Thermodynamic Quantities:
Let’s denote the key thermodynamic terms:
- qq: heat added to the system
- ww: work done by the system
- ΔU\Delta U: change in internal energy
- ΔH\Delta H: change in enthalpy
Because the process occurs quasistatically with constant external pressure and fixed temperature, the system does pressure-volume (PV) work. The work done is: w=−PextΔVw = -P_{\text{ext}} \Delta V
Where:
- Pext=2.0P_{\text{ext}} = 2.0 atm
- ΔV\Delta V: volume change due to gaseous products
Gas Volume Change:
From the balanced equation:
- Reactant gases = 1.5 mol O2O_2
- Product gases = 1 mol CO2CO_2 + 0.5 mol N2N_2 = 1.5 mol
So, no net change in moles of gas.
Thus, ΔV=0\Delta V = 0, and: w=−PextΔV=0w = -P_{\text{ext}} \Delta V = 0
But the question says the system expands and lifts a light weight, which implies some volume change. If we assume gaseous water forms instead of liquid, then: CO(NH2)2+32O2→CO2+N2+2H2O(g)\text{CO(NH}_2\text{)}_2 + \frac{3}{2} O_2 \rightarrow CO_2 + N_2 + 2 H_2O(g)
Now:
- Reactants: 1.5 mol gas
- Products: 1 (CO₂) + 1 (N₂) + 2 (H₂O) = 4 mol gas
Change in gas moles = 4−1.5=2.54 – 1.5 = 2.5 mol
Using the ideal gas law: ΔV=Δn⋅RTP\Delta V = \Delta n \cdot \frac{RT}{P} =2.5⋅0.08206⋅298.152.0≈30.6 L= 2.5 \cdot \frac{0.08206 \cdot 298.15}{2.0} \approx 30.6 \text{ L}
Now the work becomes: w=−PΔV=−2.0⋅30.6=−61.2 L\cdotpatmw = -P \Delta V = -2.0 \cdot 30.6 = -61.2 \text{ L·atm}
Convert to joules: 1 L\cdotpatm=101.325 J⇒w≈−6200 J1 \text{ L·atm} = 101.325 \text{ J} \Rightarrow w \approx -6200 \text{ J}
Conclusion:
The system does approximately 6200 J of work during the combustion. The negative sign indicates energy is lost by the system to do this work. Since temperature is constant and the piston is mobile, this is a quasistatic, isothermal expansion where chemical energy is converted into mechanical work.