A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6.0m^3. How much work is done by the gas if it expands at constant pressure to twice its initial volume? Please help ;_;
W = P(V2 - V1)
W = 1.3x10^5 (2x6 - 6 )
W = 780000 J
To find the work done by the gas when it expands at constant pressure, we can use the equation:
Work = Pressure × Change in Volume
Given:
Initial pressure, P1 = 1.3 × 10^5 Pa
Initial volume, V1 = 6.0 m^3
Final volume, V2 = 2 × V1 = 2 × 6.0 m^3 = 12 m^3
Change in Volume = Final Volume - Initial Volume
ΔV = V2 - V1 = 12 m^3 - 6.0 m^3 = 6.0 m^3
Now let's substitute the values into the formula:
Work = Pressure × Change in Volume
Work = (1.3 × 10^5 Pa) × (6.0 m^3)
Work = 7.8 × 10^5 Pa·m^3
Therefore, the work done by the gas when it expands at constant pressure to twice its initial volume is 7.8 × 10^5 Pa·m^3.
To calculate the work done by the gas, you can use the formula:
Work = Pressure × Change in Volume
Here's how you can calculate the work done by the gas in this scenario:
1. Calculate the change in volume:
To find the change in volume, subtract the initial volume from the final volume.
Final Volume = 2 × Initial Volume
Final Volume = 2 × 6.0 m^3 = 12.0 m^3
Change in Volume = Final Volume - Initial Volume
Change in Volume = 12.0 m^3 - 6.0 m^3 = 6.0 m^3
2. Substitute the given values into the formula:
Work = Pressure × Change in Volume
Work = 1.3 × 10^5 Pa × 6.0 m^3
3. Solve for the work done:
Work = 1.3 × 10^5 Pa × 6.0 m^3
Work = 7.8 × 10^5 J
Therefore, the work done by the gas as it expands at constant pressure to twice its initial volume is 7.8 × 10^5 J (joules).