A gas is compressed at a constant pressure of 0.800 atm from 6.00 L to 1.00 L. In the process, 380 J of energy leaves the gas by heat. What is the change in its internal energy?
Never mind i figured this one out.
To find the change in internal energy of the gas, we can use the first law of thermodynamics, which states that the change in internal energy (∆U) is equal to the heat transfer (Q) into or out of the system minus the work done (W) by or on the system.
Mathematically, we can represent this as:
∆U = Q - W
In this case, the problem states that 380 J of energy leaves the gas by heat (Q = -380 J). The negative sign indicates that heat is leaving the system.
Now, we need to calculate the work done (W) on the system. Since the gas is being compressed at a constant pressure, the work done can be calculated using the formula:
W = -P∆V
where P is the constant pressure and ∆V is the change in volume.
Given:
P = 0.800 atm
∆V = 1.00 L - 6.00 L = -5.00 L
Since the volume change is negative (due to compression), we need to make it positive in the equation.
∆V = -(-5.00 L) = 5.00 L
Now, we can calculate the work done:
W = -P∆V
= -(0.800 atm)(5.00 L)
= -4.00 atm∙L
With the values of Q = -380 J and W = -4.00 atm∙L, we can substitute them into the equation for ∆U:
∆U = Q - W
= (-380 J) - (-4.00 atm∙L)
= -380 J + 4.00 atm∙L
Since the units J and atm∙L are not directly compatible, we need to convert atm∙L to J using the conversion factor:
1 atm∙L = 101.3 J
Therefore:
∆U = -380 J + 4.00 atm∙L × (101.3 J/1 atm∙L)
= -380 J + 405.2 J
= 25.2 J
Hence, the change in internal energy (∆U) of the gas is 25.2 J.