A gas at a constant pressure of 5x105 Pa is cooled so that its volume decreases from 3m3 to 2.5m3. What work is performed by the gas?
IS IT LOOKING FOR INTERNAL ENERGY OR WORK???
The question is asking for the work performed by the gas. To calculate the work, we can use the formula:
Work = Pressure × Change in Volume
Given:
Pressure (P) = 5 × 10^5 Pa
Initial Volume (V₁) = 3 m³
Final Volume (V₂) = 2.5 m³
To find the change in volume (ΔV), we need to subtract the initial volume from the final volume:
ΔV = V₂ - V₁
Substituting the given values, we get:
ΔV = 2.5 m³ - 3 m³
= -0.5 m³
Note that the change in volume is negative because the gas is being compressed, resulting in a decrease in volume.
Now, we can calculate the work using the formula:
Work = Pressure × ΔV
Substituting the values, we have:
Work = (5 × 10^5 Pa) × (-0.5 m³)
= -2.5 × 10^5 J
Therefore, the work performed by the gas is -2.5 × 10^5 Joules. The negative sign indicates that work is done on the gas, as the gas is being compressed.