First Law of Thermodynamics

Question: A system absorbs 250 J of heat and does 150 J of work on the surroundings. Calculate the change in internal energy.

Solution: Given:

• Heat absorbed: $q = +250$ J (positive, endothermic)
• Work done BY system: $w = -150$ J (negative in chemistry convention)

Using First Law:

Answer: Internal energy increases by 100 J.

Question: One mole of ideal gas expands isothermally at 300 K from 2 L to 10 L. Calculate work done by the gas.

Solution: Given: $n = 1$ mol, $T = 300$ K, $V_1 = 2$ L, $V_2 = 10$ L

For isothermal process:

For isothermal ideal gas: $\Delta U = 0$ Therefore: $q = -w = +4.01$ kJ (heat absorbed)

Answer: Work done by gas = 4.01 kJ, Heat absorbed = 4.01 kJ

Question: 2 moles of a gas ($C_V = 20$ J/mol·K) is heated at constant volume from 300 K to 400 K. Calculate q, w, and ΔU.

Solution: At constant volume: $\Delta V = 0$

Work: $w = -P_{ext}\Delta V = 0$

Heat: $q_V = nC_V\Delta T = 2 \times 20 \times (400-300) = 4000$ J

Internal energy: $\Delta U = q + w = 4000 + 0 = 4000$ J

Answer: q = 4000 J, w = 0, ΔU = 4000 J

Question: A gas undergoes a cyclic process in which it absorbs 500 J of heat. What is the work done by the gas in the cycle?

Solution: For a cyclic process, the system returns to its initial state.

Therefore: $\Delta U = 0$ (state function)

Using First Law:

Work done BY the gas = 500 J (taking magnitude)

Answer: 500 J of work is done by the gas.

Key Insight: In a cyclic process, all heat absorbed is converted to work (or vice versa).

Question: One mole of ideal gas at 300 K expands from 1 atm to 0.5 atm. Calculate work done in: (a) Reversible isothermal expansion (b) Irreversible expansion against constant external pressure of 0.5 atm

Solution:

(a) Reversible isothermal:

(b) Irreversible:

For isothermal ideal gas: $PV = nRT = \text{constant}$

Initial: $P_1V_1 = 1 \times V_1$ Final: $P_2V_2 = 0.5 \times V_2$

Therefore: $V_2 = 2V_1$

But $V_1 = \frac{nRT}{P_1} = \frac{1 \times 8.314 \times 300}{1 \times 101.325} = 24.6$ L $= 0.0246$ m³

Wait, let’s use consistent units. For gases, easier to use:

Since $P_1V_1 = P_2V_2$ and $P_2 = 0.5$ atm:

Since $P_1V_1 = nRT = 1 \times 8.314 \times 300 = 2494$ J

Therefore: $V_1 = \frac{2494}{101325 \text{ Pa}} = 0.0246$ m³

Answer:

• Reversible: -1729 J (more work done BY gas)
• Irreversible: -1247 J

Key Insight: Reversible processes extract maximum work!

Question: One mole of an ideal gas ($C_V = 12.5$ J/mol·K) undergoes adiabatic expansion from 400 K to 300 K. Calculate q, w, and ΔU.

Solution:

Adiabatic process: $q = 0$ (no heat exchange)

Using First Law:

Answer: q = 0, w = -1250 J, ΔU = -1250 J

Interpretation: The gas does 1250 J of work on surroundings, and since no heat is supplied, its internal energy (and temperature) decreases.

Question: One mole of ideal gas at 300 K undergoes:

• Step 1: Isothermal expansion from 1 atm to 0.5 atm
• Step 2: Isobaric compression from 300 K to 200 K ($C_P = 29.1$ J/mol·K)

Calculate total heat absorbed and work done.

Solution:

Step 1: Isothermal expansion

For isothermal ideal gas: $\Delta U_1 = 0$

Therefore: $q_1 = -w_1 = 1729$ J

Step 2: Isobaric compression at P = 0.5 atm

(Work is positive because compression - work done ON gas)

Total:

• Total heat: $q_{total} = q_1 + q_2 = 1729 - 2910 = -1181$ J (released)
• Total work: $w_{total} = w_1 + w_2 = -1729 + 831.4 = -897.6$ J
• Total ΔU: $\Delta U_{total} = 0 + (-2078.6) = -2078.6$ J

Verification: $\Delta U_{total} = q_{total} + w_{total} = -1181 + (-897.6) = -2078.6$ J ✓

Answer: Total heat released = 1181 J, Net work done by gas = 897.6 J

Question: Two identical samples of 1 mole ideal gas at 300 K and 10 atm are expanded to 1 atm: (a) Sample A: Free expansion into vacuum (b) Sample B: Isothermal reversible expansion

For each, calculate q, w, ΔU, and final temperature.

Solution:

(a) Free expansion:

• $P_{ext} = 0$
• $w = 0$ (no opposing pressure)
• For ideal gas, $\Delta U = 0$ (depends only on T, not V)
• Therefore: $q = 0$
• Final temperature: 300 K (isothermal for ideal gas)

(b) Isothermal reversible:

• $T = 300$ K (constant)
• $\Delta U = 0$ (isothermal ideal gas)

• Final temperature: 300 K

Comparison:

Key Insight: Same initial and final states, but different paths mean different q and w values. However, ΔU is same (state function)!

Question: For real gases, explain why ΔU ≠ 0 during free expansion (Joule expansion), even though w = 0 and q = 0.

Solution:

This appears to violate the First Law, but it doesn’t!

For real gases:

• Intermolecular forces exist (van der Waals forces)
• During expansion, molecules move farther apart
• Work must be done AGAINST these attractive forces
• This work comes from internal kinetic energy
• Therefore, temperature DROPS during free expansion

Energy accounting:

• $\Delta U_{kinetic} < 0$ (molecules slow down, T decreases)
• $\Delta U_{potential} > 0$ (molecules separated against attraction)

For free expansion: $q = 0, w = 0$

Therefore: $\Delta U = 0$ still holds!

But unlike ideal gas, temperature changes even though ΔU = 0.

Real-life example: Joule-Thomson effect - gas cools when expanded through a porous plug (used in refrigeration).

For ideal gas: No intermolecular forces, so $\Delta U_{potential} = 0$, temperature remains constant.

JEE Insight: Always specify “ideal gas” when assuming ΔU = 0 for isothermal process!