Question

1. a gas expands in volume from 26.7 ml to 89.3 ml at constant temperature. calculate the work done (in joules) if the gas expands against a constant external pressure of 1.5 atm.

a. -63 j
b. -0.094 j
c. +9.5 j
d. -9.5 j
e. +4.5 j

Explanation:

Step1: Calculate the change in volume

The change in volume \(\Delta V=V_{final}-V_{initial}\).
Given \(V_{initial} = 26.7\space mL\) and \(V_{final}=89.3\space mL\), so \(\Delta V=(89.3 - 26.7)\space mL=62.6\space mL = 62.6\times10^{- 3}\space L\) (since \(1\space L = 1000\space mL\)).

Step2: Use the formula for work done against constant pressure

The formula for work done \(w=-P\Delta V\) (the negative sign is because work is done by the system). Given \(P = 1.5\space atm\).
Substitute the values: \(w=-1.5\space atm\times62.6\times 10^{-3}\space L\).
We know that \(1\space L\cdot atm=101.3\space J\).
So \(w=-1.5\times62.6\times10^{-3}\times101.3\space J\).
First, calculate \(1.5\times62.6\times10^{-3}=0.0939\).
Then \(w=-0.0939\times101.3\space J\approx - 9.5\space J\).

Answer:

D. -9.5 J