How much work (in J) is involved in a chemical reaction if the volume decreases from \(33.6 \mathrm{~L}\) to \(11.2 \mathrm{~L}\) against a constant pressure of \(90.5 \mathrm{kPa}\) ?

Short Answer

Expert verified
The work involved in the chemical reaction is \(2026.8 \mathrm{~J}\).

Step by step solution

01

Convert pressure to Pa

The pressure is given in kPa, but the SI unit for pressure is Pascal (Pa). To convert kPa to Pa, we can multiply the given pressure with 1000 since 1 kPa = 1000 Pa. So, the pressure in Pa is: 90.5 kPa × 1000 = 90500 Pa
02

Calculate the change in volume

We are given the initial volume (33.6 L) and the final volume (11.2 L). The change in volume ΔV is the difference between final and initial volumes: ΔV = V_final - V_initial ΔV = 11.2 L - 33.6 L ΔV = -22.4 L
03

Convert volume to m³

The SI unit for volume is cubic meters (m³), so we need to convert the volume ΔV from liters to m³. We can use the conversion factor that 1 L = 0.001 m³: ΔV_m³ = ΔV(L) × 0.001 ΔV_m³ = -22.4 L × 0.001 ΔV_m³ = -0.0224 m³
04

Calculate work done

Now, we can use the formula W = -PΔV to calculate the work done: W = - (90500 Pa) × (-0.0224 m³) W = 2026.8 J So, the work involved in the chemical reaction is 2026.8 J.

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