Question
Question: A sample of 0.1g of water at \({100^0}\)C and normal pressure \(\left( {1.013\times{{10}^5}N{m^{ - 2...
A sample of 0.1g of water at 1000C and normal pressure (1.013×105Nm−2) requires 54cal of heat energy of convert to steam at 1000.If the volume of the steam product is 167.1cc, the change in internal energy of the sample is?
A. 42.2J
B. 104.3J
C. 84.5J
D. 208.7J
Solution
Energy exists in many forms. For example light energy, chemical energy, heat energy. Internal energy refers to the amount of kinetic as well as potential energy possessed by the particles in a body. On the other hand, heat energy only concerns the transmission of the internal energy from hotter body to the colder body. To solve the present problem, we will deal with the thermodynamics law.
Complete step by step answer:
Thermodynamics define physical quantities. For example temperature, heat,thermodynamics work and entropy. Thermodynamics stated three laws.
First law: According to the first law when energy passes into or out of the system then internal energy of the system changes in accordance with the law of conservation of energy.
Second law: According to the second law of thermodynamics states that the total entropy of an isolated system can never decrease with time, and if all processes are reversible then it becomes constant. Isolated systems spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy or we can say that heat does not spontaneously pass from a colder body to a warmer body.
Third law: According to the third law of thermodynamics a system's entropy approaches a constant value if and only if temperature approaches absolute zero.
Given: heat energy Q is 54cal. Converted its value in joules we get Q=54×4.2j
Pressure is (1.013×105Nm−2)
Volume is 167.1cc
Now finding the change in the internal energy of the system we can use the first law of thermodynamics. According to first law:
ΔQ=ΔU+ΔW
ΔQ is a change in heat.
ΔU is a change in internal energy.
ΔW is work done.
Put the valueΔQ, pressure and volume. After putting we get
ΔU=(1.013×105)×(167.1−0.1)×10−6
We can say that the change in internal energy of the system is 208.7 J. Hence, the correct answer is option D.
Note: Always remember that gases can do the work through either expansion or compression against the constant external pressure. Thus, work done is the product of pressure and the volume during change of volume. There are usually two cases: (i) When gas does work, volume of a gas increases (ΔV>0)and work done is negative and (ii) When work is done on gas, volume of gas decreases (ΔV<0) and work done is positive.