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A gas-forming reaction produces \( 1.35 \mathrm{~m}^{3} \) of gas against a constant pressure of 179.0 kPa . Calculate the work done by the gas in joules. Hint: be sure to convert the cubic meters to liters and kilopascals to atm. Look up the conversion factors! work:

Ask by Park Bowers. in the United States
Mar 19,2025

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The work done by the gas is approximately \(-2.42 \times 10^5\) joules.

Solución

1. Convert the volume from cubic meters to liters: \[ 1.35\text{ m}^3 \times 1000\,\frac{\text{L}}{\text{m}^3} = 1350\text{ L} \] 2. Convert the pressure from kilopascals to atmospheres using the conversion factor \(1\,\text{atm} = 101.325\,\text{kPa}\): \[ 179.0\,\text{kPa} \times \frac{1\,\text{atm}}{101.325\,\text{kPa}} \approx 1.765\,\text{atm} \] 3. Calculate the work done by the gas during expansion using the formula for work at constant pressure: \[ W = -P\,\Delta V \] Here, the negative sign indicates that work is done by the system (the gas) on the surroundings. Substituting the converted values: \[ W = -\left(1.765\,\text{atm}\right)\left(1350\text{ L}\right) \] 4. Convert the work from liter·atm to joules using the conversion factor \(1\,\text{L·atm} = 101.325\,\text{J}\): \[ W = -\left(1.765 \times 1350\right)\,\text{L·atm} \times 101.325\,\frac{\text{J}}{\text{L·atm}} \] First, calculate the product in L·atm: \[ 1.765 \times 1350 \approx 2382.75\,\text{L·atm} \] Now, convert to joules: \[ W \approx -2382.75 \times 101.325\,\text{J} \approx -241550\,\text{J} \] Thus, the work done by the gas is approximately: \[ W \approx -2.42 \times 10^5\,\text{J} \]

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To calculate the work done by the gas, use the formula \( W = -P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the volume change. First, convert the pressure from kPa to atm. Using the conversion factor \( 1 \,\text{atm} = 101.325 \,\text{kPa} \), we find: \[ P = \frac{179.0 \, \text{kPa}}{101.325 \,\text{kPa/atm}} \approx 1.768 \,\text{atm} \] Next, convert the volume from cubic meters to liters. Since \( 1 \,\text{m}^3 = 1000 \,\text{L} \): \[ \Delta V = 1.35 \, \text{m}^3 \times 1000 \,\text{L/m}^3 = 1350 \,\text{L} \] Now plug in the values: \[ W = -1.768 \,\text{atm} \times 1350 \,\text{L} = -2388.8 \,\text{L} \cdot \text{atm} \] Finally, convert the work from L·atm to joules using \( 1 \,\text{L} \cdot \text{atm} = 101.325 \,\text{J} \): \[ W = -2388.8 \,\text{L} \cdot \text{atm} \times 101.325 \,\text{J/L} \cdot \text{atm} \approx -242,000 \,\text{J} \] So, the work done by the gas is approximately \( -242,000 \, \text{J} \). Did you know that gas laws have been fundamental in developing engines and even space travel? The principles used in calculating work can be found in the way rockets expel gas to generate thrust—it's all about maximizing the gas's energy and pressure for optimal propulsion! If you're looking for a practical application, consider how heating a gas in a piston can lead to work done as the gas expands! This principle is at the heart of combustion engines, where fuel is burned to create gas that pushes against pistons, converting chemical energy into mechanical work. Understanding these concepts can really enhance your grasp on how machines operate in everyday life!

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