Question

The temperature of an ideal gas having constant specific heats is given as a function of specific entropy and specific volume as \(T(s, v)=A v^{1-k} \exp \left(s / c_{v}\right)\) where \(A\) is a constant. For a reversible, constant volume process, find the expression for heat transfer per unit mass as a function of \(c_{v}\) and \(T\) using \(Q=\int T d S .\) Compare this result with that obtained by applying the first law to a closed system undergoing a constant volume process.

Short answer

Question: Compare the expressions for heat transfer in an ideal gas undergoing a constant volume process using the given temperature function \(T(s, v) = Av^{1-k}e^{s/c_v}\) and the first law of thermodynamics. Answer: The heat transfer expression for an ideal gas using the given temperature function is \(Q = A v^{1-k} \left[ c_v e^{s/c_v} \right]\), while the expression using the first law of thermodynamics is \(Q = mc_v(T_2-T_1)\). Both expressions represent the heat transfer for an ideal gas undergoing a constant volume process, but they are in different terms. The first expression is in terms of specific entropy and specific volume, while the second expression is in terms of initial and final temperatures.

Step-by-step solution

Identify the Temperature Function

The given function for the temperature of an ideal gas having constant specific heats as a function of specific entropy and specific volume is: \(T(s, v) = Av^{1-k}e^{s/c_v}\), where A is a constant.

Compute the Heat Transfer Expression

For a reversible, constant volume process, we will use the definition of heat transfer as \(Q = \int T dS\). We will rewrite the temperature function as a function of \(s\) and use it for integration: $$Q = \int T(s,v) dS = \int Av^{1-k}e^{s/c_v} dS$$ Since \(A\), \(v\) and \(c_v\) are constants and only \(s\) is variable in this case, we have: $$Q = Av^{1-k} \int e^{s/c_v} ds$$. Now integrate with respect to \(s\) and solve for \(Q\). $$Q = A v^{1-k} \left[ c_v e^{s/c_v} \right]$$

Apply first law to a closed system undergoing constant volume process

For a closed system undergoing a constant volume process, the first law of thermodynamics states: \(\Delta U = Q - W\) Since it's a constant volume process, the work done on/by the system is \(0\). Therefore, \(\Delta U = Q\) Now, for an ideal gas with constant specific heats, the change in internal energy can be expressed as: \(\Delta U = m c_v (T_2 - T_1)\), where \(m\) is the mass and \(T_1\) and \(T_2\) are the initial and final temperatures respectively. So we have: \(Q = mc_v(T_2-T_1)\)

Compare the Results

Now, let's compare the result we got from the temperature function, and integration: $$Q = A v^{1-k} \left[ c_v e^{s/c_v} \right]$$ And the result we got from applying the first law to a closed system undergoing a constant volume process: $$Q = mc_v(T_2-T_1)$$ It's not immediately apparent how these results are related, as the first expression involves specific entropy \(s\), specific volume \(v\), and the constant \(A\), while the second expression involves the mass \(m\) and the temperature change \((T_2 - T_1)\). However, both expressions represent the heat transfer for an ideal gas undergoing a constant volume process. The first expression is in terms of specific entropy and specific volume, while the second expression is in terms of initial and final temperatures. The relationship between these quantities might be more apparent if we could rewrite the temperature function in terms of initial and final states or vice versa, but as it is, both expressions represent valid ways of calculating the heat transfer for the given process.