One mole of an ideal monoatomic gas is caused to go through the cycle shown in figure. Then, the change in the internal energy is expanding the gas fr

Monjur Alahi

One mole of an ideal monoatomic gas is caused to go through the cycle shown in figure. Then, the change in the internal energy is expanding the gas from a to `c` along the path `abc` is:
A. `3P_(0)V_(0)`
B. `6 RT_(0)`
C. `4.5 RT_(0)`
D. `10.5 RT_(0)`


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Salim Ahmed Kawsar

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Correct Answer - D
`Pv= nRT` at point `C`
`2P_(0)xx4V_(0)=1xxRT_(c)`
`T_(c )=[(8P_(0)V_(0))/(R )]`
at point a
`P_(0)V_(0)=1xxRT_(0)`
`T_(0)=(P_(0)V_(0))/(R ) , T_(c )=8T_(0)`
change in internal energy `=[ nC_(v)dT]`
For path `a` to `b=1xx(3)/(2)Rxx[3T_(0))=(9)/(2)RT_(0)`
For path `b` to `c=1xx(3)/(2)Rxx[4T_(0)]=6T_(0)R`
Total change `=(9)/(2)RT_(0)+6RT_(0)=(21RT_(0))/(2)=10.5 RT_(0)`
So total change in internal energy
`Delta U =10.5 RT_(0)`

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