Physics Solution: Question 28
Analysis & Derivation
Initially, the piston is held fixed. The volume is constant. The oven supplies heat to the gas, raising its temperature at a rate \(r_0\). We can calculate the heat input power \(\dot{Q}\):
$$ \dot{Q} = n C_v \frac{dT}{dt} $$For a mono-atomic gas, \(C_v = \frac{3}{2}R\). Thus:
$$ \dot{Q} = \frac{3}{2} n R r_0 \quad \dots \text{(Equation 1)} $$When the piston is released, it is massless, meaning the forces on it are balanced at every instant (quasi-static in terms of force, though kinetic in terms of energy). The force from the gas pressure \(P\) balances the restoring force of the rubber cord \(F_{cord}\).
$$ P \cdot A = k x $$Using the Ideal Gas Law \(PV = nRT\) and noting \(V = A \cdot x\):
$$ (P \cdot A) \cdot x = n R T \implies (kx) \cdot x = n R T $$ $$ U_{gas} = \frac{3}{2}nRT = \frac{3}{2} k x^2 $$This is a crucial insight: The internal energy \(U\) depends on \(x^2\).
From the First Law of Thermodynamics (differentiated with respect to time):
$$ \dot{Q} = \frac{dU}{dt} + \frac{dW}{dt} $$We know \(\dot{Q}\) is supplied by the oven (assumed constant at the moment of release).
The work done by the gas is against the spring force: \(\frac{dW}{dt} = F v = (kx)v\).
The rate of change of internal energy is: \(\frac{dU}{dt} = \frac{d}{dt}\left(\frac{3}{2}kx^2\right) = \frac{3}{2}k(2x\dot{x}) = 3kxv\).
Substituting these into the energy balance:
$$ \dot{Q} = 3kxv + kxv = 4kxv $$Calculation
We equate the heat supply rate calculated in Phase 1 to the energy consumption rate in Phase 2 at the moment of release (where \(x = l_0\)):
$$ \frac{3}{2} n R r_0 = 4 k l_0 v $$Solving for velocity \(v\):
$$ v = \frac{3}{8} \frac{n R r_0}{k l_0} $$Substituting values:
\(n = 0.5\) mol, \(R \approx 8.314\) J/(mol K), \(r_0 = 0.08\) K/s, \(k = 25\) N/m, \(l_0 = 1.0\) m.