Unit 8: Thermodynamics — Online MCQ Test
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Practice Unit 8: Thermodynamics with a free chapter-wise online MCQ test.
This chapter covers: Thermal equilibrium and the concept of temperature zeroth lawof thermodynamics heat work and internal energy. The first lawof thermodynamics isothermal and adiabatic processes. The....
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Unit 8: Thermodynamics — Important Questions & Answers
An amount of ice of mass 10^-3 kg and temperature -10 °C is transformed to vapour of temperature 110 °C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = 2100 J/kg/K, specific heat of water = 4180 J/kg/K, specific heat of steam = 1920 J/kg/K, Latent heat of ice = 3.35 x 10^5 J/kg and Latent heat of steam = 2.25 x 10^6 J/kg)
- A. 3043 J
- B. 3024 J
- C. 3003 J
- D. 3022 J
Answer: A. 3043 J
Q1 (Ice to 0°C) = m*c_ice*deltaT = 10^-3 * 2100 * 10 = 21 J. Q2 (Melting) = m*L_f = 10^-3 * 3.35*10^5 = 335 J. Q3 (Water 0 to 100°C) = m*c_w*100 = 10^-3 * 4180 * 100 = 418 J. Q4 (Boiling) = m*L_v = 10^-3 * 2.25*10^6 = 2250 J. Q5 (Steam 100 to 110°C) = m*c_s*10 = 10^-3 * 1920 * 10 = 19.2 J. Total = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 J.
Q1 (Ice to 0°C) = m*c_ice*deltaT = 10^-3 * 2100 * 10 = 21 J. Q2 (Melting) = m*L_f = 10^-3 * 3.35*10^5 = 335 J. Q3 (Water 0 to 100°C) = m*c_w*100 = 10^-3 * 4180 * 100 = 418 J. Q4 (Boiling) = m*L_v = 10^-3 * 2.25*10^6 = 2250 J. Q5 (Steam 100 to 110°C) = m*c_s*10 = 10^-3 * 1920 * 10 = 19.2 J. Total = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 J.
Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K. If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume effect of the surrounding temperature to be negligible):
- A. 16 E
- B. E
- C. 64 E
- D. 256 E
Answer: B. E
Power P = σ A T^4. E1 = σ * (4πr1^2) * T1^4. E2 = σ * (4πr2^2) * T2^4. Substituting values: (0.8/0.2)^2 * (400/800)^4 = 16 * (1/16) = 1.
Power P = σ A T^4. E1 = σ * (4πr1^2) * T1^4. E2 = σ * (4πr2^2) * T2^4. Substituting values: (0.8/0.2)^2 * (400/800)^4 = 16 * (1/16) = 1.
A diatomic gas (gamma = 1.4) does 100 J of work when it is expanded isobarically. Then the heat given to the gas is ____ J.
- A. 250
- B. 350
- C. 400
- D. 500
Answer: B. 350
For an isobaric process, Q = n * Cp * dT and W = n * R * dT. Ratio Q/W = Cp / R = gamma / (gamma - 1). With gamma = 1.4, Q/W = 1.4 / 0.4 = 3.5. Q = 3.5 * 100 = 350 J.
For an isobaric process, Q = n * Cp * dT and W = n * R * dT. Ratio Q/W = Cp / R = gamma / (gamma - 1). With gamma = 1.4, Q/W = 1.4 / 0.4 = 3.5. Q = 3.5 * 100 = 350 J.
A diatomic gas (γ = 1.4) does 100 J of work when it is expanded isobarically. Then the heat given to the gas is ____ J.
- A. 250
- B. 300
- C. 350
- D. 400
Answer: C. 350
For an isobaric process, W = nRΔT = 100 J. Q = nCpΔT = n(f/2 + 1)RΔT. For a diatomic gas, f=5, so Cp = 7/2 R. Q = (7/2)nRΔT = 3.5 * 100 = 350 J.
For an isobaric process, W = nRΔT = 100 J. Q = nCpΔT = n(f/2 + 1)RΔT. For a diatomic gas, f=5, so Cp = 7/2 R. Q = (7/2)nRΔT = 3.5 * 100 = 350 J.
10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P1 to P2 is α Joule (P1 = 21.7 Pa and P2 = 30 Pa, Cv = 21 J/K.mol, R = 8.3 J/mol.K). The value of α is ____.
- A. 24
- B. 15
- C. 21
- D. 28
Answer: C. 21
The process is isochoric (volume constant). Heat ΔQ = n * Cv * ΔT = n * Cv * (P2 - P1) * (V/nR). Given the pressure values and Cv, calculation yields 21 J.
The process is isochoric (volume constant). Heat ΔQ = n * Cv * ΔT = n * Cv * (P2 - P1) * (V/nR). Given the pressure values and Cv, calculation yields 21 J.
Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points A and F are maintained at 100°C and 40°C respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points B and E are (close to):
- A. 89°C and 73°C respectively
- B. 80°C and 60°C respectively
- C. 80°C and 70°C respectively
- D. 60°C and 45°C respectively
Answer: A. 89°C and 73°C respectively
Assuming a series combination of conductors, the equivalent thermal resistance is calculated. Heat current H = ΔT/Req. Since conductivity Kx = 3Ky, the thermal resistance R_y = 3 * R_x. Solving the heat flow equations for junctions leads to temperatures approximately 89°C and 73°C.
Assuming a series combination of conductors, the equivalent thermal resistance is calculated. Heat current H = ΔT/Req. Since conductivity Kx = 3Ky, the thermal resistance R_y = 3 * R_x. Solving the heat flow equations for junctions leads to temperatures approximately 89°C and 73°C.
A gas undergoes a reversible change in temperature. If there is no exchange of heat in this process (ΔQ=0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
- A. CO2
- B. O2
- C. NH3
- D. He
Answer: D. He
For an adiabatic process (ΔQ=0), TV^(γ-1) = constant. Using the relation T1V1^(γ-1) = T2V2^(γ-1) with T1=T, T2=8T and V1=8V, V2=V, we get 8^(γ-1) = 8, which leads to γ-1 = 1, so γ=5/3. This value corresponds to a monoatomic gas like Helium.
For an adiabatic process (ΔQ=0), TV^(γ-1) = constant. Using the relation T1V1^(γ-1) = T2V2^(γ-1) with T1=T, T2=8T and V1=8V, V2=V, we get 8^(γ-1) = 8, which leads to γ-1 = 1, so γ=5/3. This value corresponds to a monoatomic gas like Helium.
Density of water at 4°C and 20°C are 1000 kg/m³ and 998 kg/m³ respectively. The increase in internal energy of 4 kg water when it is heated from 4°C to 20°C is ____ J. (Specific heat capacity of water = 4200 J/kg·K and 1 atmospheric pressure = 10⁵ Pa)
- A. 315826.2
- B. 234699.2
- C. 258700.8
- D. 268799.2
Answer: D. 268799.2
ΔU = Q - W. Q = mSΔT = 4 * 4200 * 16 = 268800 J. Work W = PΔV = P * m(1/ρ2 - 1/ρ1) = 10⁵ * 4 * (1/998 - 1/1000) ≈ 0.8 J. ΔU = 268800 - 0.8 = 268799.2 J.
ΔU = Q - W. Q = mSΔT = 4 * 4200 * 16 = 268800 J. Work W = PΔV = P * m(1/ρ2 - 1/ρ1) = 10⁵ * 4 * (1/998 - 1/1000) ≈ 0.8 J. ΔU = 268800 - 0.8 = 268799.2 J.
A gas is compressed adiabatically, which one of the following statement is NOT true.
- A. The temperature of the gas increases.
- B. The change in the internal energy is equal to the work done on the gas.
- C. There is no change in the internal energy
- D. There is no heat supplied to the system
Answer: C. There is no change in the internal energy
For an adiabatic process, $Q=0$. On compression, work is done on the gas, so its internal energy increases and therefore temperature generally increases for an ideal gas. Hence statements a, b, and d are true. The statement that is NOT true is that there is no change in internal energy.
For an adiabatic process, $Q=0$. On compression, work is done on the gas, so its internal energy increases and therefore temperature generally increases for an ideal gas. Hence statements a, b, and d are true. The statement that is NOT true is that there is no change in internal energy.
A flask contains Hydrogen and Argon in the ratio 2:1 by mass. The temperature of the mixture is $30^\circ\mathrm{C}$. The ratio of average kinetic energy per molecule of the two gases $(\text{Ar}/\text{K}\ \text{hydrogen})$ is (Given: Atomic Weight of Ar = 39.9)
- A. $\frac{39.9}{2}$
- B. 1
- C. 39.9
- D. 2
Answer: B. 1
For an ideal gas, average kinetic energy per molecule depends only on temperature: $$\langle K\rangle=\frac{3}{2}kT.$$ Since both gases are at the same temperature, their average kinetic energies per molecule are equal. Therefore $$\frac{\langle K\rangle_{\text{Ar}}}{\langle K\rangle_{\text{H}_2}}=1.$$ So the correct answer is 1.
For an ideal gas, average kinetic energy per molecule depends only on temperature: $$\langle K\rangle=\frac{3}{2}kT.$$ Since both gases are at the same temperature, their average kinetic energies per molecule are equal. Therefore $$\frac{\langle K\rangle_{\text{Ar}}}{\langle K\rangle_{\text{H}_2}}=1.$$ So the correct answer is 1.