Question 10.2: Determine the work done (in joules) when a sample of gas exp...
Determine the work done (in joules) when a sample of gas expands from 552 \text{mL} to 891 \text{mL} at constant temperature (a) against a constant pressure of 1.25 \text{atm}, (b) against a constant pressure of 1.00 \text{atm}, and (c) against a vacuum (1 \text{L ⋅ atm} = 101.3 \text{J}).
Strategy Determine change in volume (∆V), identify the external pressure (P), and use Equation 10.2 to calculate w. The result of Equation 10.2 will be in \text{L ⋅ atm}; use the equality 1 \text{L ⋅ atm} = 101.3 \text{J} to convert to joules.
Setup ∆V = (891 – 552)\text{mL} = 339 \text{mL}. (a) P = 1.25 \text{atm}, (b) P = 1.00 \text{atm}, (c) P = 0 \text{atm}.
w = –PΔV ^* Equation 10.2
^*Student Annotation: The \text{SI} unit of pressure is the pascal (\text{Pa}), which, in \text{SI} base units is 1 \text{kg/(m ⋅ s}^2). Volume, in \text{SI} base units is cubic meters (\text{m}^3). Therefore, multiplying units of pressure by units of volume gives [1 \text{kg/(m ⋅ s}^2)] × (\text{m}^3) = 1 ( \text{kg ⋅ m}^2)/\text{s}^2, which is the definition of the joule (\text{J}). Thus, PΔV has units of energy.
Another commonly used unit of pressure is the atmosphere (atm) [≫ Section 11.3]. Worked Example 10.2 uses atmospheres to express presure.
(a) w = –(1.25 \text{ atm})(339 \text{ mL}) (\frac{1 \text{ L}}{1000 \text{ mL}})(\frac{101.3 \text{ J}}{1 \text{ L ⋅ atm}}) = –42.9 \text{ J}
(b) w = –(1.00 \text{ atm})(339 \text{ mL}) (\frac{1 \text{ L}}{1000 \text{ mL}})(\frac{101.3 \text{ J}}{1 \text{ L ⋅ atm}}) = –34.3 \text{ J}
(c) w = –(0 \text{ atm})(339 \text{ mL}) (\frac{1 \text{ L}}{1000 \text{ mL}})(\frac{101.3 \text{ J}}{1 \text{ L ⋅ atm}}) = 0 \text{ J}