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Question

Physics Question on Thermodynamics

An ideal gas expands isothermally from a volume V1V_1 to V2 V_2and then compressed to original volume V1V_1 adiabatically. Initial pressure is p1_1 and final pressure is p3_3. The total work done is W. Then,

A

p3>p1.W>0p_3 > p_1 .W > 0

B

p3<p1.W<0p_3 < p_1 .W < 0

C

p3>p1.w<0p_3 > p_1.w < 0

D

p3p1,W0p_3 -p_1, W - 0

Answer

p3>p1.w<0p_3 > p_1.w < 0

Explanation

Solution

Slope of adiabatic process at a given state (p, F, T) is more than the slope of isothermal process. The corresponding p-V graph for the two processes is as shown in figure.
In the graph, AB is isothermal and BC is adiabatic.
WAB_{AB} = positive (as volume is increasing)
and WBCW_{BC} = negative (as volume is decreasing) plus,
WBC>WAB|W_{BC}| > |W_{AB}| as area underp- Vgraph gives the work done.
Hence, WAB_{AB} + WBC_{BC} = W < 0
From the graph itself, it is clear that p3_3 > pv_v.
Hence, the correct option is (c).
NOTE At point B, slope of adiabatic (process BC) is greater than the slope of isothermal (process AB).