Question

Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is ________ kJ mol⁻¹. (Rounded-off to the nearest integer) [Use R=8.314 J mol⁻¹K⁻¹]

Hint:

Always check the physical sensibility of a thermodynamics problem. Expansion can only occur if the internal pressure is greater than the external pressure. If the opposite is true, it's a compression. Also, pay attention to whether the requested quantity is total or molar.

Answer 1

Correct Answer: 15

Step 1: Physical interpretation A gas cannot expand against an external pressure higher than its own. Hence, the given data implies an isothermal compression from $P_i = 1.3$ MPa to $P_f = 2.1$ MPa against a constant external pressure $P_{\text{ext}} = 4.3$ MPa. Step 2: Thermodynamic relation For an ideal gas undergoing an isothermal process: \[ \Delta U = 0 \] From the first law of thermodynamics: \[ \Delta U = q + w \;\Rightarrow\; q = -w \] Step 3: Calculate initial and final volumes Using the ideal gas equation: \[ V = \frac{nRT}{P} \] Given: \[ n = 5\ \text{mol}, \quad R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}, \quad T = 293\ \text{K} \] \[ nRT = 5 \times 8.314 \times 293 = 12179\ \text{J} \] Initial volume: \[ P_i = 1.3 \times 10^6\ \text{Pa} \] \[ V_i = \frac{12179}{1.3 \times 10^6} = 9.368 \times 10^{-3}\ \text{m}^3 \] Final volume: \[ P_f = 2.1 \times 10^6\ \text{Pa} \] \[ V_f = \frac{12179}{2.1 \times 10^6} = 5.799 \times 10^{-3}\ \text{m}^3 \] Step 4: Change in volume \[ \Delta V = V_f - V_i \] \[ \Delta V = (5.799 - 9.368)\times 10^{-3} = -3.569 \times 10^{-3}\ \text{m}^3 \] (Negative sign confirms compression.) Step 5: Work done For an irreversible process at constant external pressure: \[ w = -P_{\text{ext}} \Delta V \] \[ w = -(4.3 \times 10^6)(-3.569 \times 10^{-3}) \] \[ w = +15347\ \text{J} \] Step 6: Heat transferred \[ q = -w = -15347\ \text{J} = -15.347\ \text{kJ} \] Step 7: Heat transferred per mole \[ q_{\text{molar}} = \frac{-15.347}{5} = -3.07\ \text{kJ mol}^{-1} \] Final Answer (rounded): \[ \boxed{-3\ \text{kJ mol}^{-1}} \]