What law is used to solve this?

First Law of Thermodynamics: Q = delta-U + W. Since Q (heat supplied) is same for both and delta-U is same for both, work done W must be equal for both: WA = WB.

How is work done by gas in a piston calculated?

W = P x delta-V = P x A x d where A = area of piston = pi r^2, d = displacement. At same pressure: WA = P x pi rA^2 x dA and WB = P x pi rB^2 x dB.

How is the ratio rA/rB found?

WA = WB: P pi rA^2 x 16 = P pi rB^2 x 9. rA^2/rB^2 = 9/16. rA/rB = 3/4.

What is the First Law of Thermodynamics?

Q = delta-U + W. Q = heat added to system, delta-U = change in internal energy, W = work done by system. For constant pressure process: W = P delta-V = nR delta-T. For ideal gas: delta-U = nCv delta-T.

Two gases A and B are filled at same pressure in cylinders with movable pistons of radius rA and rB. On supplying equal

Q: What is an isobaric process?

Isobaric process = constant pressure. W = P delta-V = nR delta-T. Q = nCp delta-T. delta-U = nCv delta-T. Q = delta-U + W is satisfied with Cp = Cv + R.

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PhysicsThermodynamics

Two gases A and B are filled at same pressure in cylinders with movable pistons of radius rA and rB. On supplying equal heat to both reversibly under constant pressure, pistons of A and B displace by 16 cm and 9 cm. If change in internal energy is same, then ratio rA/rB is:

Options

3/2

4/3

3/4

2/3

Correct Answer

3/4

Solution

From First Law: Q = delta-U + W

Q is same for both, delta-U is same for both → WA = WB

W = P x pi r^2 x d (same pressure, circular pistons)

P pi rA^2 x 16 = P pi rB^2 x 9

rA^2/rB^2 = 9/16

rA/rB = 3/4

First Law: Q = delta-U + W
Q same, delta-U same → WA = WB
pi rA^2 x 16 = pi rB^2 x 9 → rA/rB = 3/4

Theory: Thermodynamics

1. First Law of Thermodynamics

Q = delta-U + W. Q = heat added to system (positive when heat enters). delta-U = change in internal energy of system. W = work done BY the system (positive when system expands). Internal energy U depends only on temperature for ideal gas: delta-U = nCvT. Heat Q and work W are path functions (depend on process). Internal energy U is a state function (depends only on initial and final states).

2. Thermodynamic Processes

Isothermal (constant T): delta-U = 0, so Q = W. W = nRT ln(V2/V1). For ideal gas: PV = constant. Isobaric (constant P): W = P(V2-V1) = nR(T2-T1). Q = nCp delta-T. delta-U = nCv delta-T. Isochoric (constant V): W = 0, so Q = delta-U = nCv delta-T. Adiabatic (no heat exchange, Q = 0): delta-U = -W. PV^gamma = constant. T V^(gamma-1) = constant.

3. Work Done by Gas

General: W = integral of P dV. For constant pressure: W = P delta-V. For circular piston: delta-V = A x d = pi r^2 x d. For non-circular or varying area: must integrate. Work is the area under PV diagram. Expansion: W > 0 (gas does work on surroundings). Compression: W < 0 (surroundings do work on gas). Cyclic process: W = net area enclosed in PV diagram. Efficiency of heat engine: eta = W/Q_hot = 1 - Q_cold/Q_hot.

4. Specific Heat Capacities

Cv (constant volume): Q = nCv delta-T, all heat goes to internal energy. Cp (constant pressure): Q = nCp delta-T, heat goes to internal energy plus work. Relation: Cp - Cv = R (Mayer relation). gamma = Cp/Cv. Monatomic ideal gas (He, Ar): Cv = 3R/2, Cp = 5R/2, gamma = 5/3. Diatomic (N2, O2, air): Cv = 5R/2, Cp = 7R/2, gamma = 7/5 = 1.4. Polyatomic: gamma closer to 1. Higher gamma = more efficient adiabatic compression (diesel engines).

5. Second Law of Thermodynamics

Heat cannot spontaneously flow from cold to hot (Clausius statement). No engine can convert all heat into work with 100% efficiency (Kelvin-Planck statement). Entropy S always increases in irreversible process. dS = dQ/T for reversible process. Entropy of universe always increases. Carnot efficiency: eta_max = 1 - T_cold/T_hot. Maximum possible efficiency for any heat engine operating between these temperatures.

6. Ideal Gas Laws

Boyle law: PV = constant (constant T). Charles law: V/T = constant (constant P). Gay-Lussac law: P/T = constant (constant V). Combined: PV = nRT = NkT. n = moles, R = 8.314 J/mol/K, N = number of molecules, k = Boltzmann constant = 1.38x10^-23 J/K. Real gas deviations: van der Waals equation (a/V^2 - repulsion + pressure, b - volume correction). At low T and high P, real gas deviates most from ideal behaviour.

7. Kinetic Theory of Gases

Pressure = (1/3) rho v_rms^2 = (1/3)(m N/V) v_rms^2. RMS speed v_rms = sqrt(3RT/M) = sqrt(3kT/m). Average speed v_avg = sqrt(8RT/pi M). Most probable speed v_p = sqrt(2RT/M). Ratio: v_p : v_avg : v_rms = 1 : 1.128 : 1.225. Mean free path lambda = 1/(sqrt(2) pi d^2 n). Average kinetic energy per molecule = (3/2)kT. For ideal gas: internal energy U = (f/2)nRT where f = degrees of freedom.

8. Heat Engines and Refrigerators

Heat engine: absorbs Q_H from hot reservoir, does work W, rejects Q_C to cold reservoir. Efficiency eta = W/Q_H = 1 - Q_C/Q_H. Carnot engine (reversible): eta = 1 - T_C/T_H (maximum possible). Refrigerator: reverse of heat engine. Work input W to move heat Q_C from cold to hot. COP = Q_C/W. Heat pump: COP = Q_H/W = 1 + COP_refrigerator. Carnot COP = T_C/(T_H - T_C). Real engines are less efficient than Carnot due to irreversibilities (friction, heat leaks, rapid pressure changes).

Frequently Asked Questions

1. Why must the work done by both gases be equal? ⌄

From First Law: Q = delta-U + W. Rearranging: W = Q - delta-U. Given: Q is equal for both gases (equal heat supplied), delta-U is equal for both (given in problem). Therefore W = Q - delta-U is equal for both. This is a direct consequence of the First Law — if two quantities are equal (Q and delta-U), the third (W) must also be equal.

2. Why does the problem specify constant pressure? ⌄

At constant pressure, work W = P x delta-V = P x A x displacement. Since pressure is constant and the same for both gases, we can factor it out. The circular piston area is A = pi r^2. So work WA = P x pi rA^2 x 16 and WB = P x pi rB^2 x 9. Setting WA = WB allows us to find the ratio rA/rB. If pressure were not constant (or different), we could not use this simple formula.

3. What is an isobaric process? ⌄

Isobaric process = constant pressure. In this problem, both gases expand at constant pressure (movable pistons with same external pressure on both). Work done W = P delta-V. Heat Q = nCp delta-T. Change in internal energy delta-U = nCv delta-T. Since Q = nCp delta-T and W = nR delta-T, we have Q/W = Cp/R = Cp/(Cp-Cv) = 1/(1-1/gamma). The process is fully characterized by pressure remaining constant while volume changes.

4. What would rA/rB be if displacements were equal instead? ⌄

If displacements were equal (say both = d), then: WA = WB gives pi rA^2 x d = pi rB^2 x d, so rA = rB. The ratios would be 1:1. The different displacements (16 and 9) are what create the different radii. The piston with larger radius must displace less to do the same work (larger area compensates for smaller displacement). Ratio: rA^2/rB^2 = dB/dA = 9/16, giving rA/rB = 3/4.

5. What is Mayer relation Cp - Cv = R and why? ⌄

At constant volume: all heat goes to internal energy, Q_V = nCv delta-T. At constant pressure: heat goes to internal energy PLUS work of expansion, Q_P = nCp delta-T = nCv delta-T + P delta-V = nCv delta-T + nR delta-T. Dividing by n delta-T: Cp = Cv + R. The extra R accounts for the work done pushing the surroundings at constant pressure. This R comes from the energy needed for the gas to expand. Cp is always greater than Cv by exactly R for ideal gas.

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