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A system performs work ΔW when an amount of heat is ΔQ added to the system, the corresponding change in the internal energy is ΔU. A unique function of the initial and final states (irrespective of the mode of change) is -
Change in internal energy (ΔU) depends upon initial an find state of the function while ΔQ and ΔW are path dependent also.
A container of volume 1m3 is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at 300 K. The other compartment is vaccum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be -
This is the case of free expansion and in this case , so temperature remains same i.e. 300 K.
110 J of heat is added to a gaseous system, whose internal energy change is 40 J, then the amount of external work done is
⇒
Which of the following is not thermodynamical function ?
Work done is not a thermodynamical function.
When the amount of work done is 333 cal and change in internal energy is 167 cal, then the heat supplied is -
First law thermodynamics states that
Heat always refers to energy in transit from one body to another because of temperature difference.
A thermo-dynamical system is changed from state (P1, V1) to (P2, V2) by two different process. The quantity which will remain same will be
Change in internal energy does not depend upon path so remain constant.
If 150 J of heat is added to a system and the work done by the system is 110 J, then change in internal energy will be
If ΔQ and ΔW represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as
∵ Heat supplied to the system so ΔQ → Positive
and work is done on the system so ΔW → Negative
Hence +ΔQ = ΔU – ΔW
Which of the following can not determine the state of a thermodynamic system
State of a thermodynamic state cannot determine by a single variable (P or V or T)
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