Gas in a container is at a pressure of 1.6 atm and a volume of 6.0 m3.(a) What is the work done on the gas if it expands at constant pressure to twice its initial volume? J(b) What is the work done on the gas if it is compressed at constant pressure to one-quarter of its initial volume? J
Question
Gas in a container is at a pressure of 1.6 atm and a volume of 6.0 m3.(a) What is the work done on the gas if it expands at constant pressure to twice its initial volume? J(b) What is the work done on the gas if it is compressed at constant pressure to one-quarter of its initial volume? J
Solution
(a) The work done on a gas at constant pressure can be calculated using the formula:
W = P * ΔV
where: W is the work done, P is the pressure, and ΔV is the change in volume.
In this case, the gas is expanding to twice its initial volume, so the change in volume is:
ΔV = V_final - V_initial = 2V - V = V
Substituting the given values:
W = P * ΔV = 1.6 atm * 6.0 m^3 = 9.6 atm*m^3
However, we need to convert this to Joules. The conversion factor is 101.3 J/(atm*m^3), so:
W = 9.6 atmm^3 * 101.3 J/(atmm^3) = 972.48 J
So, the work done on the gas as it expands is 972.48 Joules.
(b) If the gas is compressed to one-quarter of its initial volume, the change in volume is:
ΔV = V_final - V_initial = 0.25V - V = -0.75V
The work done is:
W = P * ΔV = 1.6 atm * -0.75 * 6.0 m^3 = -7.2 atm*m^3
Converting to Joules:
W = -7.2 atmm^3 * 101.3 J/(atmm^3) = -729.36 J
So, the work done on the gas as it is compressed is -729.36 Joules. The negative sign indicates that work is done on the gas (as opposed to by the gas).
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