A gas is contained in a cylinder with a pressure of 120kPa and is initial volume of 0.66. How much work is done by the gas as it expands at constant pressure to twice its volume, or is compressed to one-third its initial volume?
1 Answer
Notice how it gives you a pressure, an initial volume, a final volume, and notes that it is all at constant pressure. Also, assume ideality here. You are in one of those "movable-piston" scenarios.
This question is asking you about the fundamental equation for expansion/compression work, which is:
#\mathbf(delw = -PdV)#
If you integrate this, you get:
#int delw = color(green)(w) = -int_(V_1)^(V_2) PdV = color(green)(-P(V_2 - V_1))#
Next, we should consider what it means to expand or compress the gas.
- When you expand the gas, the gas does the work. Hence, work should turn out to be numerically negative (
#V_2 - V_1 > 0# , hence#w < 0# ). - When you compress the gas, you do work on the gas. Hence, work should turn out to be numerically positive (
#V_2 - V_1 < 0# , hence#w > 0# ).
1) If we expand the gas at constant-pressure (isobaric) conditions to twice its initial volume, we get:
#w_"exp" = -int_(V_1)^(V_2) PdV#
#= -P int_("0.66 L")^("1.32 L")dV#
#= -("120000" cancel"Pa" xx ("1 bar")/(10^5 cancel"Pa"))(1.32 - "0.66 L")#
#= -"0.792 L"*"bar"#
Now to convert this into
#-"0.792"cancel("L"cdot"bar") xx ("8.314472 J")/("0.083145" cancel("L"*"bar"))#
#color(blue)(w_"exp") ~~# #color(blue)(-"79.1997 J")#
2) This second part is very similar. Just remember that work is now positive due to doing work on the gas instead of allowing the gas to do work.
#w_"comp" = -int_(V_1)^(V_2) PdV#
#= -P int_("0.66 L")^("0.22 L")dV#
#= -("120000" cancel"Pa" xx ("1 bar")/(10^5 cancel"Pa"))(0.22 - "0.66 L")#
#= "0.528 L"*"bar"#
And the final conversion:
#"0.528" cancel("L"cdot"bar") xx ("8.314472 J")/("0.083145" cancel("L"*"bar"))#
#color(blue)(w_"comp") ~~# #color(blue)("52.7998 J")#
