T(K) = T(ºC) = 273; T must be in K
1 cal = 4.18 J
R = 8.314 J/mol•K = 0.0821 L•atm/mol•K;
8.314 J = .0821 L•atm
Constant
P
V
T
PV work
-P∆V
0 no work possible
-nRT ln(Vf/Vin)
PV = nRT
∆E = nRT
∆E = (qin – qout) + w
__________________________________________________________________________
=PV = nRT=∆E =(qin – qout) + w= PV = nRT=∆E =(qin – qout) + w= PV = nRT=∆E =(qin – qout) + w= PV = nRT=∆E
How many situations can we solve for the heat exchanged, the change in internal energy
and the work done? In all the following, we will assume no leaks (no change in n).
Constant P : We then have V1/T1 = V2/T2
We know how to calculate the work, if we
know the change in volume.
We can calculate the change in internal
energy if we know the change in
temperature.
We can calculate the heat by difference.
w = -P∆V
∆E = nR∆T
∆E = (qin – qout) + w
We must know one of n or P and three of V1 , T1 , V2 , T2 or q.
Examples:
[1] A 3 L sample of gas at 1.2 atm and 300K is cooled to 250K at constant pressure. Calculate the
work done, the heat flow and the change in internal energy.
From PV = nRT, n = PV/RT = (1.2 atm)(3 L)/(.0821 L•atm/mol•K)(300K)
= 0.146 mol
V1/T1 = V2/T2 gives (3 L)/(300 K) = V2/(250 K); V2 = 2.5 L
w = -P∆V = - (1.2 atm) (2.5 L – 3.0 L) = 0.6 L•atm = 60.8 J
∆E = nR∆T = (0.146 mol)(8.314 J/mol•K)(250 K – 300 K) = -60.7 J
qnet = ∆E – w = -60.7 J - 60.8 J = -121.5 J
[2] 109.3 Joules of heat are added to 1.5 L of gas at 1.1 atm. and 290 K. The final volume of the gas
is 2.0 L at 1.1 atm. Calculate the work and change in internal energy.
V1/T1 = V2/T2 gives (1.5 L)/(290 K) = (2.0 L)/T2 T2 = 387 K
From PV = nRT, n = PV/RT = (1.1 atm)(1.5 L)/(.0821 L•atm/mol•K)(290K)
= 0.0693 mol
w = -P∆V = -(1.1 atm)(2.0-1.5L) = -.55 L•atm. = -55.7J
∆E = qnet + w = 109.3J -55.7J = 53.6 J
or, we could have used ∆E = nR∆T = (0.0693 mol)(8.314 J/mol•K)(387-290K)
= 53.58J and we didn’t need n.
Constant V : We then have P1/T1 = P2/T2
The work = 0
We can calculate the change in internal
energy if we know the change in
temperature.
We can calculate the heat from the change
in internal energy
w=0
∆E = nR∆T
∆E = (qin – qout) + w = (qin – qout)
We need to know n and either the net heat gain or ∆T
Examples:
[3] A sealed, rigid container holding 0.2 moles of gas exerts 1.5 atm of pressure at 300K. The
container is warmed to 350K
Calculate the change in internal energy and the heat added to the gas.
∆E = nR∆T = (0.2 mol)(8.314 J/mol•K)(350 – 300K) = 83.14 J
since w = 0, ∆E = q = 83.14 J
We could also determine the final pressure :P1/T1 = P2/T2
which gives (1.5 atm)/(300 K) = P2/(350K) P2 = 1.75 atm
And PV = nRT allows us to calculate teh volume, but we don’t need these.
[4] A 3 L container holds gas at 350K and a pressure of 0.75 atm. The gas is cooled until the
pressure reaches 0.60 atm.
Calculate the change in internal energy and the heat added to the gas.
P1/T1 = P2/T2 gives (0.75 atm)/(350 K) = (0.60 atm)/ T2 T2 = 280K
∆E = nR∆T so we need n :
n = PV/RT = (0.75 atm)(3L)/(.0821 L•atm/mol•K)(350K) = 0.078 mol
∆E = nR∆T = (0.078 mol)(8.314 J/mol•K)(280-350K) = -45.39 J
∆E = q = -45.39 J (heat flowed from the gas)
Constant T : We then have P1V1 = P2V2
The work = nRT ln(V2/V1)
The change in internal energy = 0 since
∆T=0
We can calculate the heat from the work
w = -nRT ln(V2/V1)
∆E = 0
∆E = 0; (qin – qout) = nRT ln(V2/V1)
We need to know n, T, V1 and V2.
Examples:
[5] A container with 0.5 moles of gas at a pressure of 1.5 atm and a volume of 2.73 L expands
isothermally to a volume of 4.09 L. Calculate the work done and the heat gained by the gas.
We need the temperature, so use PV = nRT :
T = PV/nR = (1.5 atm)(2.73 L)/(0.5 mol)( .0821 L•atm/mol•K) = 99.76 K
Now, w = -nRT ln(V2/V1) = -(0.5 mol)( .0821 L•atm/mol•K)(99.76 K)ln(4.09/2.73)
= -1.655 L•atm = -168 J (work done by the gas)
heat : q = -w = 168 J
[6] 4.0 L of gas at 300K and 0.75 atm are heated at constant temperature until the pressure is
1.0 atm. Calculate the work done and the heat gained by the gas.
We need both n and the final volume. Use PV = nRT to find n and final V.
n = PV/RT = (0.75 atm)(4.0L)/(0.0821 L•atm/mol•K)(300L) = 0.122 mol
P1V1 = P2V2 gives (0.75 atm)(4.0L) = (1.00 atm)V2 V2 = 3.0L
w = -nRT ln(V2/V1) = -(0.122 mol)(8.314 J/mol•K)(300K) ln(3.0L/4.0L)
= -87.54 J (gas did work on surroundings) so q = -w = 87.54 J
[7] Using the following graph of the process, calculate the
work done during the cycle
A to B to C to A,
The PV work is equal to the area under the curve on
the PV graph.
The process CA contributes zero work.
The process BC contributes P1(V1-V2) or negative the
area under the BC line.
The work for process AB is the area under the curve
AB.
The total work done during the cycle is equal to the area under the AB curve minuss
the area under the line BC. In other words, it is the area enclosed by the cycle!