A gas expands against a constant external pressure of 1.7 bar until its volume has increased from 5.4 L to 8.9 L. During this process it absorbs 640.0 J of heat from the surroundings. Calculate the energy change of the gas.
delta U = Heat in - Work out = Q - (P*delta V)
= 640 - 1.7*101.3*10^5 N/m^2*3.5*10^-3 m^3
= ? Joules
It looks like it will be a negative number
To calculate the energy change of the gas, we need to use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
In this case, the gas absorbs 640.0 J of heat from the surroundings, so the heat added to the system is 640.0 J.
To find the work done by the system, we can use the equation:
Work = Pressure * Change in Volume
Given that the external pressure is constant at 1.7 bar and the volume has increased from 5.4 L to 8.9 L, the change in volume is:
Change in Volume = Final Volume - Initial Volume
= 8.9 L - 5.4 L
= 3.5 L
Now we can calculate the work done by the system:
Work = Pressure * Change in Volume
= 1.7 bar * 3.5 L
= 5.95 bar·L
Since we're using SI units, we need to convert the pressure from bar to pascal (Pa). 1 bar is equal to 100,000 Pa.
Work = 5.95 bar·L * 100,000 Pa/bar
= 595,000 J
Now we can calculate the energy change of the gas using the first law of thermodynamics:
Energy Change = Heat Added - Work Done
= 640.0 J - 595,000 J
= -594,360 J
Therefore, the energy change of the gas is approximately -594,360 J.