Question

Calculate the PV type of work for the following reaction at 1 bar pressure. C3H6(g) + HCl(g) → C3H7Cl(g)}

• A. 5.2 J
• B. 10.21 J
• C. 15.00 J
• D. 18.2 J

Hint:

Remember that work done by expansion is negative and work done by compression is positive. Use the conversion factor 1 dm³ bar = 100 J.

Answer 1

The Correct Option is A

extbf{Step 1: } Identify the formula for PV work. The work done by or on a system due to a change in volume against an external pressure is given by: \[ W = -P_{ext} \Delta V \] where $W$ is the work, $P_{ext}$ is the external pressure, and $\Delta V$ is the change in volume ($V_{final} - V_{initial}$). extbf{Step 2: } Determine the initial volume ($V_1$) and the final volume ($V_2$). The reaction involves gaseous reactants and a gaseous product. The initial volume is the sum of the volumes of the reactants: \[ V_1 = V_{C_3H_6} + V_{HCl} \] The final volume is the volume of the product: \[ V_2 = V_{C_3H_7Cl} \] Given volumes are: \[ V_{C_3H_6} = 150 \text{ mL} \] \[ V_{HCl} = 150 \text{ mL} \] \[ V_{C_3H_7Cl} = 150 \text{ mL} \] extbf{Step 3: } Calculate the initial volume and final volume: \[ V_1 = 150 \text{ mL} + 150 \text{ mL} = 300 \text{ mL} \] \[ V_2 = 150 \text{ mL} \] extbf{Step 4: } Calculate the change in volume ($\Delta V$): \[ \Delta V = V_2 - V_1 = 150 \text{ mL} - 300 \text{ mL} = -150 \text{ mL} \] extbf{Step 5: } Convert the volumes to a consistent unit for pressure-volume work. It is common to use cubic decimeters (dm³) for volume when pressure is in bar, as 1 dm³ bar = 100 J. \[ 1 \text{ mL} = 1 \text{ cm}^3 \] \[ 1 \text{ dm}^3 = 1000 \text{ cm}^3 = 1000 \text{ mL} \] So, to convert mL to dm³: divide by 1000. \[ \Delta V = -150 \text{ mL} = -150 / 1000 \text{ dm}^3 = -0.150 \text{ dm}^3 \] extbf{Step 6: } Use the given external pressure ($P_{ext}$). The external pressure is given as 1 bar. \[ P_{ext} = 1 \text{ bar} \] extbf{Step 7: } Calculate the work done using the formula from Step 1: \[ W = -P_{ext} \Delta V = -(1 \text{ bar}) \times (-0.150 \text{ dm}^3) \] \[ W = 0.150 \text{ dm}^3 \text{ bar} \] extbf{Step 8: } Convert the work from dm³ bar to Joules (J). The conversion factor is 1 dm³ bar = 100 J. \[ W = 0.150 \text{ dm}^3 \text{ bar} \times 100 \text{ J/dm}^3 \text{ bar} \] \[ W = 15.0 \text{ J} \] extbf{Step 9: } Interpret the sign of the work. A positive value for work means that work is done *by* the system on the surroundings (expansion work). In this case, the system contracts, so work is done *on* the system by the surroundings.