A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages of expansion/compression the states of gases are not defined). The work done can be calculated using
dw = – P ext dV
while in case of reversible process the work done can be calculated using
dw = – PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process. Since P = 
, so
w = 
= – 
. dV = –nRT ln 
Since dw = – PdV so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram.
(i) An ideal gaseous sample at initial state i (P 0 , V 0 ,T 0 ) is allowed to expand to volume 2V 0 using two different process; in the first process the equation of process is PV 2 = K 1 and in second process the equation of the process is PV = K 2 . Then,
Text Solution
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(i) Work done in isothermal process will be more than PV 2 = const, process whatever be the value of K 1 and K 2 as is shown in the diagram.
(ii) Clearly (V f ) isothermal > (V f ) adiabatic
(iii) Work done in isothermal process is more than in adiabatic process as shown in the diagram above
(iv) For isothermal irreversible process (T = constant)
P° V° = P ´ 2V° 
P = 
For P 2 V = constant
V° = P 2 2V° 
P = 
Final pressure of isothermal process > Final pressure of adiabatic process.
and Final pressure of irreversible adiabatic process > Final pressure of reversible adiabatic process.
So, final pressure order II > I > IV > III.
(v) P°,V° 
2V°, 
For reversible adiabatic,
(2V°) γ = P 1 V 0 γ 
P 1 = P°2 γ –1
For reversible PV 2 = K
(2V°) 2 = P 2 V 0 2 
P 2 = 2P°
So, P 2 > P 1
Since final volume is same
P 
T
So, T 2 > T 1
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