a) What is meant by entropy ? Write the first law of thermodynamics in terms of entropy. [1+1] b) Can a room be cooled by leaving the door of an electric refrigerator open ? Justify. [1] c) A…

a)

Entropy (S) is a thermodynamic property that measures the degree of disorder or randomness in a system. It quantifies how energy is distributed among the microscopic particles of a system. The SI unit of entropy is J·K⁻¹. Entropy increases when a system becomes more disordered (e.g., melting of ice, mixing of gases) and decreases when it becomes more ordered (e.g., freezing of water).

The first law of thermodynamics in terms of entropy states that for a reversible process, the change in entropy ($dS$) of a system is given by: $$dS=\frac{d{Q}_{\text{rev}}}{T}$$ where:

• $d{Q}_{\text{rev}}$ = infinitesimal heat transfer in a reversible process,
• $T$ = absolute temperature of the system.

For an irreversible process, the entropy change is: $$\Delta S\ge \int \frac{dQ}{T}$$ where the equality holds only for reversible processes.

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b)

No, a room cannot be cooled by leaving the door of an electric refrigerator open. Here’s why:

1. Heat Rejection: A refrigerator works by transferring heat from its inside (cooler region) to the outside (warmer room). When the door is open, the heat absorbed from the room’s air is rejected back into the same room.
2. Energy Consumption: The refrigerator’s compressor consumes electrical energy, which is converted into heat during operation. This heat is also released into the room.
3. Net Effect: The total heat released into the room (from both the absorbed heat and the electrical energy converted to heat) is greater than the heat removed from the room. Thus, the room warms up instead of cooling down.

Conclusion: Leaving the refrigerator door open increases the room’s temperature and wastes electricity.

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c)

Given:

• Initial temperature, $T_1=30^{\circ }\mathrm{C}=303\mathrm{K}$ (converted to Kelvin),
• Final volume, $V_2=\left(\frac{8}{27}\right)V_1$,
• Ratio of specific heats, $\gamma =\frac{C_p}{C_v}=\frac{5}{3}$ (for a monoatomic gas, $\gamma =\frac{5}{3}$ is correct).

Assumption: The process is adiabatic (no heat exchange with surroundings, $Q=0$).

Step 1: Relate volume and temperature for an adiabatic process

For an adiabatic process involving an ideal gas: $$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$$ Rearranged to solve for $T_2$: $$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$$

Step 2: Substitute given values

$$\gamma -1=\frac{5}{3}-1=\frac{2}{3}$$ $$\frac{V_1}{V_2}=\frac{V_1}{\left(\frac{8}{27}\right)V_1}=\frac{27}{8}$$ $$T_2=303\times {\left(\frac{27}{8}\right)}^{\frac{2}{3}}$$

Step 3: Simplify the exponent

$${\left(\frac{27}{8}\right)}^{\frac{2}{3}}={\left({\left(\frac{27}{8}\right)}^{\frac{1}{3}}\right)}^2={\left(\frac{3}{2}\right)}^2=\frac{9}{4}=2.25$$

Step 4: Calculate $T_2$

$$T_2=303\times 2.25=684.75\mathrm{K}$$

Step 5: Find the change in temperature ($\Delta T$)

$$\Delta T=T_2-T_1=684.75-303=381.75\mathrm{K}$$ Convert back to Celsius (if required, though the question asks for change in temperature, which is the same in Kelvin and Celsius): $$\Delta T=381.75^{\circ }\mathrm{C}$$

Final Answer: The change in temperature is $381.75\mathrm{K}$ (or $381.75^{\circ }\mathrm{C}$). The temperature increases by this amount.