1.00kj of energy transferred as heat to O2, external pressure is 2.00atm. O2 expands from 1.00L to 3.00L against this constant pressure. Calculate w and change in internal energy (delta U) when O2 is ideal gas?
Suppose that 1.00 kJ of energy is transferred as heat to oxygen in a cylinder fitted with a piston; the external pressure is 2.00 atm. The oxygen expands from 1.00 L to 3.00 L against this constant pressure. Calculate w and delta U for the entire process by treating the O2 as an ideal gas.
Suppose that 1.00 kJ of energy is transferred as heat to oxygen in a cylinder fitted with a piston; the external pressure is 2.00 atm. The oxygen expands from 1.00 L to 3.00 L against this constant pressure. Calculate w and delta U for the entire process by treating the O2 as an ideal gas.
1 Answer
#DeltaU = "0.595 kJ"#
Well, if the gas is ideal, we assume it doesn't interact within itself, so the expansion is supposedly reversible.
From the first law of thermodynamics,
#DeltaU = q + w#
#= q - int_(V_1)^(V_2) PdV# where
#q# is heat flow with respect to the system,#w# is work done by or on the system,#P# is the EXTERNAL pressure, and#V# is the volume.
The heat flow is already given:
#q = ul(+"1.00 kJ")#
Heat was transferred INTO
The work done at a constant external pressure is:
#w = -P int_(V_1)^(V_2)dV#
#= -P DeltaV#
#= -"2.00 atm" cdot ("3.00 L" - "1.00 L")#
#= -"4.00 L"cdot"atm"#
This SHOULD be negative. The system expands, doing work on the surroundings.
Do the units match up? (No!) Should they? (They had better.)
#-4.00 cancel("L"cdot"atm") xx (8.314472 cancel"J")/(0.082057 cancel("L"cdot"atm")) xx ("1 kJ")/(1000 cancel"J")#
#= ul(-"0.405 kJ")#
Therefore, the change in internal energy is:
#color(blue)(DeltaU) = "1.00 kJ" + (-"0.405 kJ") = color(blue)("0.595 kJ")#
