As its volume is doubled from an initial value of 0.088 m3, a sample of gas is held at a constant pressure of 3.4 × 104 Pa by heating. Then, while a constant volume is maintained, the gas is cooled back to its initial temperature. The work that the gas does on the environment during the entire process is most nearly
6.0 × 103 J
3.0 × 103 J
−3.0 × 103 J
−6.0 × 103 J
Well, that's quite a gas-tastic question! Let's see if we can clown around with some calculations here.
Since the pressure is constant, we can use the equation: work = pressure × change in volume. Since the volume is doubling, the change in volume is V_final - V_initial = 2V_initial - V_initial = V_initial.
Therefore, the work done on the environment during the entire process is given by work = pressure x change in volume = 3.4 × 10^4 Pa x 0.088 m^3 = 2.992 × 10^3 J.
So the answer is most nearly 3.0 x 10^3 J. Ta-da!
To find the work done by the gas, we can use the equation:
Work = Pressure * Change in Volume
First, let's calculate the change in volume:
Change in Volume = Final Volume - Initial Volume
Given that the volume is doubled, the final volume is:
Final Volume = 2 * Initial Volume = 2 * 0.088 m^3 = 0.176 m^3
The change in volume is:
Change in Volume = 0.176 m^3 - 0.088 m^3 = 0.088 m^3
Now, we can calculate the work done:
Work = Pressure * Change in Volume
Work = 3.4 × 10^4 Pa * 0.088 m^3
Work = 2.992 × 10^3 J
Therefore, the closest answer is 3.0 × 10^3 J.
To find the work done by the gas on the environment during the entire process, we can use the formula:
Work = Pressure * Change in Volume
In this case, the pressure is constant at 3.4 × 10^4 Pa. Let's calculate the change in volume first.
The volume of the gas doubles from its initial value of 0.088 m^3. Doubling the initial volume gives us 2 * 0.088 m^3 = 0.176 m^3.
So, the change in volume is 0.176 m^3 - 0.088 m^3 = 0.088 m^3.
Now, we can calculate the work done by the gas on the environment using the formula:
Work = 3.4 × 10^4 Pa * 0.088 m^3
Work ≈ 2.992 × 10^3 J
The work done by the gas on the environment during the entire process is approximately 2.992 × 10^3 J, which is closest to 3.0 × 10^3 J.
Therefore, the answer is 3.0 × 10^3 J.