A one mole sample of an ideal monatomic gas at 400 K undergoes an isovolumetric process acquring 550 J of energy by heat. Next it undergoes an isobaric process during which is loses the same amount of energy by heat. Find the work done on the gas. Assume three degrees of freedom.
To find the work done on 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.
First, let's calculate the change in internal energy of the gas during the isovolumetric process.
The isovolumetric process is a constant volume process, which means the gas does not change its volume. In this case, no work is done because work (W) is given by:
W = PΔV
where P is the pressure and ΔV is the change in volume. Since ΔV is zero in an isovolumetric process, the work done (W) is zero.
Now, let's calculate the change in internal energy (ΔU) during the isovolumetric process. The change in internal energy is given by:
ΔU = Q - W
where Q is the heat added to the system.
In this case, the gas acquires 550 J of energy by heat during the isovolumetric process, so Q = 550 J. Since W = 0, the change in internal energy (ΔU) is equal to the heat added to the system:
ΔU = 550 J
Next, let's calculate the change in internal energy (ΔU') during the isobaric process.
The isobaric process is a constant pressure process, which means the pressure of the gas remains constant. In this case, the work done (W') is given by:
W' = PΔV
where P is the pressure and ΔV is the change in volume.
Since the amount of energy lost by heat during the isobaric process is the same as the energy gained during the isovolumetric process (550 J), we know that Q' = -550 J.
Using the equation ΔU' = Q' - W', we can substitute the values:
ΔU' = -550 J - W'
Since the change in internal energy (ΔU') is equal to the heat added (Q') minus the work done (W'), we can rearrange the equation to solve for W':
W' = -550 J - ΔU'
Now, we need to calculate the change in internal energy (ΔU'). The change in internal energy for a monatomic gas is given by:
ΔU' = (3/2) nR ΔT
where n is the number of moles, R is the gas constant, and ΔT is the change in temperature.
In this case, we have a one mole sample of gas, so n = 1. The gas constant R is approximately 8.314 J/(mol·K). The temperature changes from 400 K to an unknown value.
ΔU' = (3/2) * (1) * (8.314 J/(mol·K)) * ΔT
ΔU' = 12.471 J/mol·K * ΔT
Now, we substitute this value back into the equation for W':
W' = -550 J - 12.471 J/mol·K * ΔT
To find the value of ΔT, we need to use the ideal gas law. The ideal gas law is given by:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the absolute temperature.
Since the process is isobaric (constant pressure), the pressure remains constant. Therefore:
PΔV = nRΔT
Since the volume is constant in the isovolumetric process, ΔV is equal to zero. Therefore:
0 = nRΔT
From this, we find that ΔT = 0.
Now we can substitute ΔT = 0 into the equation for W':
W' = -550 J - 12.471 J/mol·K * ΔT
W' = -550 J - 12.471 J/mol·K * 0
W' = -550 J
Therefore, the work done on the gas during the isobaric process is -550 J.