Question

a sample of oxygen contracts in volume from 7.57 to 3.13 l at a constant temperature. calculate the work done, in j, if the gas contracts due to a constant pressure of 1.09 atm. report your answer to the nearest whole number.
note that 1 atm = 101.325 kpa.

Explanation:

Step1: Convert pressure to Pa

Given \(1\ atm = 101.325\ kPa=101325\ Pa\), and pressure \(P = 1.09\ atm\).
So \(P=1.09\times101325\ Pa\).

$$P = 1.09\times101325=110444.25\ Pa$$

Step2: Calculate the change in volume

Initial volume \(V_1 = 7.57\ L\), final volume \(V_2=3.13\ L\).
\(\Delta V=V_2 - V_1=(3.13 - 7.57)\ L=- 4.44\ L\).
Convert volume to \(m^3\) (since \(1\ m^3 = 1000\ L\)), \(\Delta V=-4.44\times10^{- 3}\ m^3\).

Step3: Calculate work done

The formula for work done at constant pressure is \(w = P\Delta V\).
Substitute \(P = 110444.25\ Pa\) and \(\Delta V=-4.44\times10^{-3}\ m^3\) into the formula.

$$w=110444.25\times(-4.44\times10^{-3})$$

$$w=-490.37247\ J\approx - 490\ J$$

Answer:

\(-490\)