Question

Wall of the container are adiabatic and monatomic gas is filled in both compartment. Piston is movable and non-conducting. As heater is switch on, gas in the left chamber expand pushing the piston until pressure become $\frac{243}{32}P_0$, both side, Find i)T, V of each इब्बा. e. (i) $\Delta U$ for each इब्बा (iii) (WD)$_{gas}$ ,, ,, (iv) Heat supply by heater

Answer 1

i) Left: $V_{fL} = \frac{46}{27}V_0$, $T_{fL} = \frac{207}{16}T_0$; Right: $V_{fR} = \frac{8}{27}V_0$, $T_{fR} = \frac{9}{4}T_0$ ii) Left: $\Delta U_L = \frac{573}{32}P_0V_0$; Right: $\Delta U_R = \frac{15}{8}P_0V_0$ iii) $W_L = \frac{\frac{191}{16}P_0V_0}{1-k}$, where $k = \frac{\ln(243/32)}{\ln(27/46)}$ iv) $Q_L = \frac{573}{32}P_0V_0 + \frac{\frac{191}{16}P_0V_0}{1-k}$

Answer 2

The problem involves a thermodynamic process in a container divided by a movable, non-conducting piston. The walls of the container are adiabatic. Both compartments are filled with a monatomic gas. Heat is supplied to the left compartment, causing it to expand and push the piston until equilibrium is reached at a final pressure $P_f = \frac{243}{32}P_0$.

Assumptions:

1. Both compartments contain the same number of moles of monatomic gas, $n$.
2. The initial state of both compartments is $P_{0L} = P_{0R} = P_0$, $V_{0L} = V_{0R} = V_0$, and $T_{0L} = T_{0R} = T_0$.
3. For a monatomic gas, the adiabatic index is $\gamma = \frac{C_p}{C_v} = \frac{5}{3}$.

Analysis of the Right Compartment (R): The right compartment is sealed by adiabatic walls and a non-conducting piston, so the process is adiabatic ($Q_R = 0$). Initial state: $P_{0R} = P_0$, $V_{0R} = V_0$, $T_{0R} = T_0$. Final state: $P_{fR} = P_f = \frac{243}{32}P_0$, $V_{fR}$, $T_{fR}$.

For an adiabatic process, $P V^\gamma = \text{constant}$. $P_0 V_0^{5/3} = \frac{243}{32}P_0 V_{fR}^{5/3}$ $V_{fR}^{5/3} = \frac{32}{243} V_0^{5/3} = \left(\frac{2^5}{3^5}\right)^{3/5} V_0^{5/3} = \left(\frac{2}{3}\right)^3 V_0^{5/3}$ $V_{fR} = \frac{8}{27} V_0$.

For an adiabatic process, $T P^{(1-\gamma)/\gamma} = \text{constant}$. $T_0 P_0^{(1-5/3)/(5/3)} = T_{fR} P_f^{(1-5/3)/(5/3)}$ $T_0 P_0^{-2/5} = T_{fR} \left(\frac{243}{32}P_0\right)^{-2/5}$ $T_{fR} = T_0 \left(\frac{243}{32}\right)^{2/5} = T_0 \left(\frac{3^5}{2^5}\right)^{2/5} = T_0 \left(\frac{3}{2}\right)^2 = \frac{9}{4} T_0$.

Analysis of the Left Compartment (L): Heat is supplied to the left compartment ($Q_L > 0$). The gas expands, doing work ($W_L > 0$). Initial state: $P_{0L} = P_0$, $V_{0L} = V_0$, $T_{0L} = T_0$. Final state: $P_{fL} = P_f = \frac{243}{32}P_0$, $V_{fL}$, $T_{fL}$.

Assuming the total volume of the container is conserved: $V_{fL} + V_{fR} = V_{0L} + V_{0R} = 2V_0$. $V_{fL} = 2V_0 - V_{fR} = 2V_0 - \frac{8}{27}V_0 = \frac{46}{27}V_0$.

Using the ideal gas law ($PV=nRT$): $T_{fL} = \frac{P_{fL} V_{fL}}{nR} = \frac{(\frac{243}{32}P_0) (\frac{46}{27}V_0)}{P_0 V_0 / T_0} = \frac{243}{32} \times \frac{46}{27} \times T_0 = \frac{9 \times 46}{32} T_0 = \frac{207}{16} T_0$.

i) T, V of each compartment:

• Left Compartment (L): $V_{fL} = \frac{46}{27}V_0$, $T_{fL} = \frac{207}{16}T_0$
• Right Compartment (R): $V_{fR} = \frac{8}{27}V_0$, $T_{fR} = \frac{9}{4}T_0$

ii) $\Delta U$ for each compartment: For a monatomic gas, $\Delta U = n C_v \Delta T = n \frac{3}{2} R (T_f - T_i)$.

• Left Compartment (L): $\Delta U_L = n \frac{3}{2} R \left(\frac{207}{16}T_0 - T_0\right) = n \frac{3}{2} R \left(\frac{191}{16}\right)T_0 = \frac{573}{32} nRT_0 = \frac{573}{32} P_0V_0$.
• Right Compartment (R): $\Delta U_R = n \frac{3}{2} R \left(\frac{9}{4}T_0 - T_0\right) = n \frac{3}{2} R \left(\frac{5}{4}\right)T_0 = \frac{15}{8} nRT_0 = \frac{15}{8} P_0V_0$.

iii) (WD)$_{gas}$ (Work done by the gas in the left chamber): The process in the left chamber is a polytropic process $P_L V_L^k = \text{constant}$. $P_0 V_0^k = P_f V_{fL}^k \implies \frac{P_f}{P_0} = \left(\frac{V_0}{V_{fL}}\right)^k \implies \frac{243}{32} = \left(\frac{27}{46}\right)^k$. $k = \frac{\ln(243/32)}{\ln(27/46)}$. The work done is $W_L = \frac{P_{fL}V_{fL} - P_{0L}V_{0L}}{1-k}$. $P_{fL}V_{fL} = nRT_{fL} = \frac{207}{16} nRT_0 = \frac{207}{16} P_0V_0$. $W_L = \frac{\frac{207}{16}P_0V_0 - P_0V_0}{1-k} = \frac{\frac{191}{16}P_0V_0}{1-k}$.

iv) Heat supply by heater: Using the First Law of Thermodynamics for the left chamber: $Q_L = \Delta U_L + W_L$. $Q_L = \frac{573}{32} P_0V_0 + \frac{\frac{191}{16}P_0V_0}{1-k}$.