Adiabatic Compression in a Composite System
Problem Overview: We have a system with two chambers separated by a movable partition. We are calculating the work done by an external agent to compress the entire system adiabatically. Since the system is thermally isolated ($Q=0$), the First Law of Thermodynamics tells us:
$$W_{\text{external}} + W_{\text{atmosphere}} = \Delta U_{\text{system}}$$
Step 1: Adiabatic Compression of Right Chamber
The external force pushes the piston in. The gas in the right chamber ($m_2$) is compressed until its pressure rises to match the left chamber ($P_L$). Let the new volume of the right chamber be $V’$.
Using the adiabatic relation $PV^\gamma = \text{constant}$ and the Ideal Gas Law $P = \frac{mRT}{MV}$: $$ P_R V_0^\gamma = P_L (V’)^\gamma $$ $$ \left( \frac{m_2 R T_i}{M V_0} \right) V_0^\gamma = \left( \frac{m_1 R T_i}{M V_0} \right) (V’)^\gamma $$
Simplifying yields the volume ratio: $$ \frac{V’}{V_0} = \left( \frac{m_2}{m_1} \right)^{1/\gamma} $$
The temperature $T’$ of the right chamber just before mixing is found using $TV^{\gamma-1} = \text{const}$: $$ T’ = T_i \left( \frac{V_0}{V’} \right)^{\gamma-1} = T_i \left[ \left( \frac{m_1}{m_2} \right)^{1/\gamma} \right]^{\gamma-1} = T_i \left( \frac{m_1}{m_2} \right)^{\frac{\gamma-1}{\gamma}} $$
Step 2: Mixing of Gases
The valve opens and the gases mix. Internal energy is conserved ($U_{\text{total}} = U_1 + U_2$).
$$ (m_1 + m_2) c_v T_{\text{mix}} = m_1 c_v T_i + m_2 c_v T’ $$ $$ T_{\text{mix}} = \frac{m_1 T_i + m_2 T’}{m_1 + m_2} $$
Step 3: Final Compression and Full Algebra
The piston continues to push until the total volume is reduced to $V_0$. The process is adiabatic from the mixed state volume ($V_0 + V’$) to $V_0$.
$$ T_f = T_{\text{mix}} \left( \frac{V_0 + V’}{V_0} \right)^{\gamma-1} = T_{\text{mix}} \left( 1 + \frac{V’}{V_0} \right)^{\gamma-1} $$
Detailed Algebraic Derivation of $T_f$
Substitute $T’ = T_i (m_1/m_2)^{\frac{\gamma-1}{\gamma}}$ into the mixing equation: $$ T_{\text{mix}} = \frac{m_1 T_i}{m_1 + m_2} \left[ 1 + \frac{m_2}{m_1} \frac{T’}{T_i} \right] $$ $$ T_{\text{mix}} = \frac{m_1 T_i}{m_1 + m_2} \left[ 1 + \frac{m_2}{m_1} \left( \frac{m_1}{m_2} \right)^{\frac{\gamma-1}{\gamma}} \right] $$ Using exponent rules ($x^1 \cdot x^{-a} = x^{1-a}$): $$ \frac{m_2}{m_1} \left( \frac{m_2}{m_1} \right)^{-\frac{\gamma-1}{\gamma}} = \left( \frac{m_2}{m_1} \right)^{1 – \frac{\gamma-1}{\gamma}} = \left( \frac{m_2}{m_1} \right)^{1/\gamma} $$ So, $$ T_{\text{mix}} = \frac{m_1 T_i}{m_1 + m_2} \underbrace{\left[ 1 + \left( \frac{m_2}{m_1} \right)^{1/\gamma} \right]}_{\text{Term X}} $$
We know $\frac{V’}{V_0} = (\frac{m_2}{m_1})^{1/\gamma}$. $$ \left( 1 + \frac{V’}{V_0} \right)^{\gamma-1} = \underbrace{\left[ 1 + \left( \frac{m_2}{m_1} \right)^{1/\gamma} \right]^{\gamma-1}}_{\text{Term } X^{\gamma-1}} $$
$$ T_f = T_{\text{mix}} \times (\text{Compression Factor}) $$ $$ T_f = \left( \frac{m_1 T_i}{m_1 + m_2} \cdot X^1 \right) \cdot X^{\gamma-1} $$ $$ T_f = \frac{m_1 T_i}{m_1 + m_2} \cdot X^{1 + \gamma – 1} = \frac{m_1 T_i}{m_1 + m_2} \cdot X^\gamma $$ Result: $$ T_f = T_i \frac{m_1}{m_1 + m_2} \left[ 1 + \left( \frac{m_2}{m_1} \right)^{1/\gamma} \right]^\gamma $$
Step 4: Calculation of Work Done
Using the First Law: $W_{\text{ext}} = \Delta U – W_{\text{atm}}$.
- $\Delta U = (m_1 + m_2) c_v (T_f – T_i)$
- $W_{\text{atm}} = -P_0 \Delta V_{\text{piston}} = -P_0 (A l_i)$ (Work done by atmosphere is positive work on gas, so we subtract it from total energy need)
Numerical Substitution
Given: $m_1 = 12\text{g}, m_2 = 2\text{g}$, $\gamma=5/3$ (He), $T_i=273.15\text{K}$, $P_0=10^5\text{Pa}$, $V_0 = 1.12 \times 10^{-2} \text{m}^3$.
1. Calculate ratio term: $\left( \frac{2}{12} \right)^{0.6} \approx 0.341$. Bracket term: $[1.341]^{5/3} \approx 1.628$.
2. Calculate Mass Fraction: $\frac{12}{14} \approx 0.857$.
3. Final Temperature: $T_f = 273.15 \times 0.857 \times 1.628 \approx 381.1 \text{ K}$.
Change in Internal Energy ($\Delta U$):
$$ \Delta U = (0.014 \text{ kg}) (3150 \text{ J/kgK}) (381.1 – 273.15) $$
$$ \Delta U = 44.1 \times 107.95 \approx 4760 \text{ J} $$
Atmospheric Work ($W_{\text{atm}}$):
$$ P_0 V_0 = 10^5 \times (0.01 \times 1.12) = 1120 \text{ J} $$
External Work: $$ W_{\text{ext}} = 4760 – 1120 = 3640 \text{ J} $$