ThermodynamicsHard
Question
For an ideal gas
Options
A.the change in internal energy in a constant pressure process from temperature T1 to T2 is equal to nCv (T2 - T1) where Cv is the moalr heat capacity at constant volume and n the number of moles of the gas
B.the change in internal energy of the gas and the work done by the gas are equal in magnitude in an diabatic process.
C.the change in internal energy of the gas and the work done by the gas are equal in magnitude in an diabetic process.
D.no heat is added or removed in an adiabatic process
Solution
(a) ᐃU = nCv ᐃT = nCv (T2 - T2) in all process
(b) In adiabatic process ᐃQ = 0
∴ ᐃU = - ᐃW or |ᐃU| = |ᐃW|
(c) In isothermal process ᐃT = 0 ∴ ᐃU = 0
(d) In adiabatic process ᐃQ = 0
∴ All the options are correct.
(b) In adiabatic process ᐃQ = 0
∴ ᐃU = - ᐃW or |ᐃU| = |ᐃW|
(c) In isothermal process ᐃT = 0 ∴ ᐃU = 0
(d) In adiabatic process ᐃQ = 0
∴ All the options are correct.
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
A black body, at a temperature of 727oC, radiates heat at a rate of 20 cal m-2s-1. When its temperature is raised to 727...The unit of gas constant R is :-...In the diagrams (i) to (iv) of variation of volume with changing pressure is shown. A gas is taken along the path ABCDA....One mole of an ideal gas in initial state A undergoes a cyclic processABCA, as shown in the figure. Its pressure at A is...The molar specific heat under constant pressure of oxygen is CP = 7.03 cal/mole k. The quantity of heat required to rais...