Choose the best possible answer.
- Assume we can change the equilibrium state of a system via two different processes. Assume that the initial and final states are the same. Which of the quantities Δu, ΔQ, ΔW and ΔT must be the same for the two processes?
(a)Only ΔQ and ΔW
(b) Only Δu and ΔT c. Only ΔQ and ΔT d. Only Δu and ΔW
Understand the question. The state of the system is changed by two different processes. However, the final states attained in both processes is the same. This means the state functions had the same values in the final states. Now four quantities are given and it is asked that which of them would have the same values in both cases. Remember only the state functions would have the same value because the state functions do not depend on the path followed. Here internal energy and temperature are the state functions and heat and work are NOT. Therefore, only internal energy and temperature would have the same value for the two processes.
- In any process the maximum amount of mechanical energy that can be converted to heat,
(A). Depends upon the amount of friction
(B). Depends upon the intake and exhaust temperatures
(C). Depends upon whether kinetic or potential energy is involved.
(C). Is 100%
- In isothermal change, internal energy
||c. Becomes zero
||d. Remains constant
Temperature ∝ Internal energy.
- A thermo bottle containing hot coffee is vigorously shaken. Consider coffee is the system, its temperature
||b. Decreases below 0°C
||c. Remains the same
By shaking the bottle, we do work on it. It increases the internal energy of the system.
- Maximum work can be obtained in the process called
By first law of thermodynamics, when heat is given to a system, it partly increases the internal energy of the system and partly is used in doing work. In an isothermal process, the change in internal energy is zero and all the heat provided is converted to work done. Hence, W is maximum.
- A heat engine takes in 800 J of heat at 1000 K and exhausts 600 J of heat at 400 K. What is the actual efficiency of that engine?
Efficiency = η = (Q1
. Put values, η = (800 – 600)/800 = 200/800 = 0.25. Percent efficiency = 0.25 × 100 = 25%.)
- If the temperature of the heat source is increased, the efficiency of Carnot engine
||c. Remains same
||d. First increases then becomes constant
: η = (T1
.) where T1
is the temperature of the source.
- Triple point of water is
|a. 273.16° F
||b. 372.16 K
||d. 273.16 K
- A real gas can be approximated to an ideal gas at
|a. Low density
||b. High pressure
||c. High density
||d. Low temperature
A real gas behaves closely to an ideal gas at high temperature and low pressure. When temperature is high and pressure is low, the gas has larger volume and low density.
- If the volume of the gas is to be increased by 4 times, then
Explanation: The relation in pressure, temperature and volume is PV = RT ⇒ V = RT/P ⇒ V ∝ T/P.
- Temperature and pressure must be doubled
- At constant P the temperature must be increased by four times.
- At constant T the pressure must be increased by four times.
- It cannot be increased.
Option A: When temperature and pressure is doubled, the volume suffers no change. Because T is directly and P is indirectly proportional to V and both increases in P and T will cancel out.
Option 2: When P is constant and temperature increases four times, this will also increase the volume four fold. Because both volume and temperature are directly proportional. Correct option.
Option 3: Prove incorrectness by looking at the proportionality above.
Option 4: Prove incorrectness by looking at the proportionality above.
Explanation: Entropy is increase of disorder. When the gas cools, its molecular motion slows down and probably, a disorder is changed to an orderly fashion. However, the disorder increases in all the remaining three processes.
- In which of the systems listed below entropy decreasing.
(A) A gas is cooled
(B) A plate is shattered.
(C) An egg is scrambled
(D) A drop of dye diffuses in a cup of water
- If the temperature of the source and sink of a Carnot engine having efficiency η are each decreased by 100 K, then the efficiency η
Explanation: Efficiency of a Carnot engine is η = 1 – (T2/T1). Now consider the ratio T2/T1. If there is an equal decrease in the numerator and denominator, the overall ratio decreases. This increases the algebraic sentence 1 – (T2/T1). Thus increasing the efficiency. Take and example,
Let, T1 = 200 K and T2 = 150 K, then T2/T1 = 150/200 = ¾ ⇒ 1- ¾ = ¼. Now let there is a decrease of 100 in both values, then T2 = 50 K and T1 = 100 K, therefore, T2/T1 = 50/100 = ½ and 1 – (T2/T1) = 1 – ½ = ½ which has increased.
|a. Remains constant
||b. Becomes 1