Question

a sample of nitrogen expands in volume from 1.1 to 6.3 l at a constant temperature. calculate the work done, in j, if the gas expands against a constant pressure of 1.25 atm. report your answer to the nearest whole number. note that 1 atm = 101.325 kpa.

Explanation:

Step1: Convert pressure to Pa

Given \(P = 1.25\ atm\). Since \(1\ atm=101325\ Pa\), then \(P = 1.25\times101325\ Pa\).

$$P = 126656.25\ Pa$$

Step2: Calculate the change in volume

\(\Delta V=V_{final}-V_{initial}\). Given \(V_{initial} = 1.1\ L\) and \(V_{final}=6.3\ L\). Since \(1\ L = 10^{- 3}\ m^{3}\), \(\Delta V=(6.3 - 1.1)\times10^{-3}\ m^{3}\)

$$ \Delta V=5.2\times10^{-3}\ m^{3}$$

Step3: Use the formula for work done at constant pressure

The formula for work done \(W=-P\Delta V\) (the negative sign indicates work done by the gas). Substitute \(P = 126656.25\ Pa\) and \(\Delta V = 5.2\times10^{-3}\ m^{3}\)

$$W=-126656.25\times5.2\times10^{-3}$$

$$W=- 658.6125\ J$$

The magnitude of work done (work done by the gas) is \(|W| = 659\ J\)

Answer:

\(659\)