In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are C_v = 3 \over 2R , C_p = 5 \over 2R , and those for an ideal diatomic gas are C_v = 5 \over 2R , C_p = 7 \over 2R . Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

Heat and Thermodynamics PYQ — JEE Advanced 2014 · Shift 2 · Q55 with Solution

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Correct answer: D

Understand the process

The whole container (including piston and walls) is perfectly insulating, so no heat is exchanged with outside.
The partition between the two gases is thermally conducting, so heat can flow between the gases.
The partition is now free to move without friction, so in equilibrium both gases will have:
- same temperature,
- same pressure.

Since the partition is movable and conducting, the two gases together behave like an isolated system internally redistributing energy.

Final equilibrium temperature

Because the whole system is thermally insulated from outside, total internal energy is conserved except for work done by the system through the outer piston.

But to compute the final common temperature first, note that the movable partition only redistributes energy internally, and at final equilibrium both gases have the same temperature $T_f$ .

Initial internal energies:

Monatomic gas: $n_1=2$ , $T_1=700\,\text{K}$ , $C_{v1}=\frac{3}{2}R$
Diatomic gas: $n_2=2$ , $T_2=400\,\text{K}$ , $C_{v2}=\frac{5}{2}R$

Since ideal-gas internal energy depends only on temperature, $U = n C_v T$

At equilibrium, if common temperature is $T_f$ , then from energy balance between the two gases:

= 2\left(\frac{3}{2}R\right)T_f + 2\left(\frac{5}{2}R\right)T_f + W_{\text{ext correction?}}$$ However, since the system as a whole can do work through the outer piston, a cleaner route is to use the fact that the external pressure on the piston remains equal to the common gas pressure, so the relevant conserved quantity for the two-gas system under adiabatic evolution is obtained most simply by using enthalpy balance for final common-pressure state. For the combined insulated system with only boundary work, $$\Delta U = -W$$ and since finally both gases have the same pressure and temperature, the total work done equals the decrease in total enthalpy divided appropriately. An easier standard way here is: For each gas evolving quasi-statically to the same final pressure and temperature, total work done by both gases is $$W = -\Delta U_{\text{total}}$$ so we first need $T_f$. Because the process is slow and the partition is conducting and movable, the final state is the same as that obtained by equating total enthalpy at common pressure in an adiabatically isolated composite system: $$n_1 C_{p1} T_1 + n_2 C_{p2} T_2 = (n_1 C_{p1} + n_2 C_{p2})T_f$$ Thus, $$2\left(\frac{5}{2}R\right)(700) + 2\left(\frac{7}{2}R\right)(400) = \left[2\left(\frac{5}{2}R\right) + 2\left(\frac{7}{2}R\right)\right]T_f$$ $$5R\cdot 700 + 7R\cdot 400 = (5R+7R)T_f$$ $$3500R + 2800R = 12R\,T_f$$ $$6300R = 12R\,T_f$$ $$T_f = 525\,\text{K}$$ 3. **Compute change in total internal energy** Initial total internal energy: $$U_i = 2\left(\frac{3}{2}R\right)(700) + 2\left(\frac{5}{2}R\right)(400)$$ $$U_i = 3R\cdot 700 + 5R\cdot 400 = 2100R + 2000R = 4100R$$ Final total internal energy: $$U_f = 2\left(\frac{3}{2}R\right)(525) + 2\left(\frac{5}{2}R\right)(525)$$ $$U_f = 3R\cdot 525 + 5R\cdot 525 = 8R\cdot 525 = 4200R$$ So, $$\Delta U = U_f - U_i = 4200R - 4100R = 100R$$ 4. **Use first law for the whole system** For the combined two-gas system, no heat enters from outside: $$Q=0$$ Hence, $$\Delta U = Q - W = -W$$ Therefore, $$W = -\Delta U = -100R$$ So the **total work done by the gases** is $$\boxed{-100R}$$ 5. **Check options** - A: $250R$ ❌ - B: $200R$ ❌ - C: $100R$ ❌ - D: $-100R$ ✅ Therefore the correct option is **D**.

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View written solutionFree

Correct answer: D

Understand the process

The whole container (including piston and walls) is perfectly insulating, so no heat is exchanged with outside.
The partition between the two gases is thermally conducting, so heat can flow between the gases.
The partition is now free to move without friction, so in equilibrium both gases will have:
- same temperature,
- same pressure.

Since the partition is movable and conducting, the two gases together behave like an isolated system internally redistributing energy.

Final equilibrium temperature

Because the whole system is thermally insulated from outside, total internal energy is conserved except for work done by the system through the outer piston.

But to compute the final common temperature first, note that the movable partition only redistributes energy internally, and at final equilibrium both gases have the same temperature $T_f$ .

Initial internal energies:

Monatomic gas: $n_1=2$ , $T_1=700\,\text{K}$ , $C_{v1}=\frac{3}{2}R$
Diatomic gas: $n_2=2$ , $T_2=400\,\text{K}$ , $C_{v2}=\frac{5}{2}R$

Since ideal-gas internal energy depends only on temperature, $U = n C_v T$

At equilibrium, if common temperature is $T_f$ , then from energy balance between the two gases:

= 2\left(\frac{3}{2}R\right)T_f + 2\left(\frac{5}{2}R\right)T_f + W_{\text{ext correction?}}$$ However, since the system as a whole can do work through the outer piston, a cleaner route is to use the fact that the external pressure on the piston remains equal to the common gas pressure, so the relevant conserved quantity for the two-gas system under adiabatic evolution is obtained most simply by using enthalpy balance for final common-pressure state. For the combined insulated system with only boundary work, $$\Delta U = -W$$ and since finally both gases have the same pressure and temperature, the total work done equals the decrease in total enthalpy divided appropriately. An easier standard way here is: For each gas evolving quasi-statically to the same final pressure and temperature, total work done by both gases is $$W = -\Delta U_{\text{total}}$$ so we first need $T_f$. Because the process is slow and the partition is conducting and movable, the final state is the same as that obtained by equating total enthalpy at common pressure in an adiabatically isolated composite system: $$n_1 C_{p1} T_1 + n_2 C_{p2} T_2 = (n_1 C_{p1} + n_2 C_{p2})T_f$$ Thus, $$2\left(\frac{5}{2}R\right)(700) + 2\left(\frac{7}{2}R\right)(400) = \left[2\left(\frac{5}{2}R\right) + 2\left(\frac{7}{2}R\right)\right]T_f$$ $$5R\cdot 700 + 7R\cdot 400 = (5R+7R)T_f$$ $$3500R + 2800R = 12R\,T_f$$ $$6300R = 12R\,T_f$$ $$T_f = 525\,\text{K}$$ 3. **Compute change in total internal energy** Initial total internal energy: $$U_i = 2\left(\frac{3}{2}R\right)(700) + 2\left(\frac{5}{2}R\right)(400)$$ $$U_i = 3R\cdot 700 + 5R\cdot 400 = 2100R + 2000R = 4100R$$ Final total internal energy: $$U_f = 2\left(\frac{3}{2}R\right)(525) + 2\left(\frac{5}{2}R\right)(525)$$ $$U_f = 3R\cdot 525 + 5R\cdot 525 = 8R\cdot 525 = 4200R$$ So, $$\Delta U = U_f - U_i = 4200R - 4100R = 100R$$ 4. **Use first law for the whole system** For the combined two-gas system, no heat enters from outside: $$Q=0$$ Hence, $$\Delta U = Q - W = -W$$ Therefore, $$W = -\Delta U = -100R$$ So the **total work done by the gases** is $$\boxed{-100R}$$ 5. **Check options** - A: $250R$ ❌ - B: $200R$ ❌ - C: $100R$ ❌ - D: $-100R$ ✅ Therefore the correct option is **D**.

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