Analyzing the Process Relationship
The given process for an ideal gas is defined by the relationship \(P = kT\), where \(P\) is pressure, \(T\) is temperature, and \(k\) is a constant.
We use the ideal gas law: \(PV = nRT\) where \(V\) is volume, \(n\) is the number of moles, and \(R\) is the universal gas constant.
Substitute the given process condition (\(P = kT\)) into the ideal gas law:
\( (kT)V = nRT \)Dividing both sides by \(kT\) (assuming \(T \neq 0\) and \(k \neq 0\)), we get:
\( V = \frac{nR}{k} \)Since \(n\), \(R\), and \(k\) are constants, this equation shows that the volume \(V\) remains constant throughout the process.
Determining Molar Heat Capacity
A process where volume is constant is known as an isochoric process.
The molar heat capacity (\(C\)) for any process is defined as:
\( C = \frac{dQ}{n \, dT} \)where \(dQ\) is the heat added to the gas.
According to the first law of thermodynamics:
\( dQ = dU + dW \)where \(dU\) is the change in internal energy and \(dW\) is the work done by the gas.
For an ideal gas, the change in internal energy depends only on temperature:
\( dU = n C_V dT \)where \(C_V\) is the molar heat capacity at constant volume.
The work done by the gas is given by:
\( dW = P \, dV \)Since the volume is constant in this process (\(V = \text{constant}\)), the change in volume \(dV = 0\). Therefore, the work done is:
\( dW = P \times 0 = 0 \)Substituting \(dU\) and \(dW\) into the first law:
\( dQ = n C_V dT + 0 \) \( dQ = n C_V dT \)Now, substitute this expression for \(dQ\) into the definition of molar heat capacity \(C\):
\( C = \frac{n C_V dT}{n \, dT} \)Simplifying the expression gives:
\( C = C_V \)Thus, for an ideal gas undergoing the process \(P = kT\), the molar heat capacity \(C\) is equal to the molar heat capacity at constant volume (\(C_V\)).
Similar Questions
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