A monoatomic ideal gas has Cp= 5R/2
a) four moles of the gas undergoes a process in which the temperature of the gas increases from 300 K to 500 K while the pressure is kept constant at 2.00 x 105 Pa. What is the change in the internal energy of the gas?
b) Four moles of the gas undergoes a process for which the volume is 3.00 m3 and is kept constant while the pressure decreases from 4500 Pa to 3600 Pa. What is the heat flow Q for this process?
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To solve these problems, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat flow (Q) into the system minus the work done (W) by the system.
The general equation for the first law of thermodynamics is:
ΔU = Q - W
a) In this case, the process is described as being at constant pressure. When the pressure is constant, the work done by the gas can be represented by the equation:
W = PΔV
where P is the constant pressure and ΔV is the change in volume.
However, since the volume is not given in this problem, we cannot directly calculate the work done. However, we do have the specific heat capacity at a constant pressure (Cp) for a monoatomic ideal gas, which is given as Cp = 5R/2.
The specific heat capacity at constant pressure is related to the change in internal energy by the equation:
ΔU = nCpΔT
where n is the number of moles, Cp is the specific heat capacity at constant pressure, and ΔT is the change in temperature.
Since we know the values of n (4 moles), Cp (5R/2), and ΔT (500 K - 300 K = 200 K), we can calculate the change in internal energy using the equation:
ΔU = nCpΔT
= (4 mol)(5R/2)(200 K)
= 2000 R mol K
Therefore, the change in internal energy of the gas is 2000 R mol K.
b) In this case, the process is described as being at constant volume. When the volume is constant, the work done by the gas is zero (W = 0).
The heat flow (Q) for this process can be calculated using the equation:
Q = ΔU + W
Since there is no work done (W = 0), the heat flow becomes:
Q = ΔU
Using the same equation as in part a), we can calculate the change in internal energy:
ΔU = nCpΔT
= (4 mol)(5R/2)(0 K)
= 0 J
Therefore, the heat flow (Q) for this process is 0 J.