If W is the work done by a system against its surrounding, what will –W stand for ?
If W is work done by the system then –W would be the work done on the system by its surroundings.
The correct answer is: Work done on the system by its surroundings
Internal energy of a perfect gas depends on.
dU = nCvdT.
The correct answer is: temperature only
In a cyclic process.
Because the process is cyclic, there is no change in internal energy after each cycle. Therefore the net work done in each cycle equals the heat added to the system.
Step 2 - Isochoric process: The work done is W2 = 0. Heat Q2 is removed from the system because the temperature decreases from T1 to T2.
Which of the following is not a property of the system?
Heat is thermal energy transferred from a hotter body to a cooler body that arein contact. Hence, it is not a property of the system.
The correct answer is: Heat
If a system A is in thermal equilibrium separately with B and C, then B and C are also in thermal equilibrium with each other. This is the statement for.
The above statement is nothing but zeroth law of thermodynamics.
The correct answer is: Zeroth law of thermodynamics
The processes or systems that do not involve heat is called.
In adiabatic processes, dQ = 0 heat remains fixed in the process, hence adiabatic processes do not involve heat.
The correct answer is: Adiabatic processes
Let ΔWi = amount of work done by the gas when compressed isothermally to volume V, ΔQ = amount of heat absorbed by the gas during the process, ΔWa = amount of work done by the gas when expanded adiabatically to volume V, ΔE = change in internal energy of the gas due to complete process. So ΔE is given by.
dU = dQ – dW
⇒ ΔU = ΔQ – ΔW
ΔW, work done by gas when compressed isothermally is positive.
dU = dQ – (ΔdWi-ΔWa)
Hence the correct answer would be ΔQ- ΔWi-ΔWa
First law of the thermodynamics is conservation of.
dθ = dU + dW
The correct answer is: energy
The piston containing an ideal gas is originally is the state x (see figure). The gas is taken through a thermal cycle as shown.
The work done by the gas is positive, if the direction of the thermal cycle is
Since work done = Area between p-V curve and volume axis
Hence, work in positive direction is clockwise
Wxy = Area xyBA
Wyx = Area yBAx
The correct answer is: clockwise
In the pV diagram below, which is the correct statement?
In a cyclic process as shown above, the area enclosed by the curve gives the work done by the system.
The correct answer is: The area abcda represents the work done by the system for the process