An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature `T_1`, pressure `P_1` and volume `V_1`and the spring is in its relaxed state. The gas is then heated very slowly to temperature `T_2`,pressure `P_2`and volume `V_2`. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement (s) is (are)

Options

A. if `V_(2)=2V_(1)` and `T_(2)=3T_(1)` , then the energy in the spring is `(1)/(4)P_(1)V_(1)`
B. if `V_(2)=2V_(1)` and `T_(2)=3T_(1)` , then the change in internal energy is `3P_(1)V_(1)`
C. if `V_(2)=3V_(1)` and `T_(2)=4T_(1)` , then the work done by the gas is `(7)/(3) P_(1)V_(1)`
D. if `V_(2)=3V_(1)` and `T_(2)=4T_(1)` , then the heat supplied to the gas is `(17)/(6) P_(1)V_(1)`

Answer

A, B, C

Explanation

`P_(i)=P_(1)` In final position of pistion, froce balance gives
`Kx =(P_(2)-P_(1))A`
Also,`(P_(2)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
If `V_(2)=2V_(1),T_(2)=3T_(1),` then
`P_(2)=(3)/(2)P_(1), Also , V_(2)-V_(1)=Ax`
so , `kx=(P_(2)-P_(1))((V_(2)-V_(1)))/(x)`
`(1)/(2)Kx^(2)=(1)/(2)(P_(2)-P_(1))(V_(2)-V_(1))=(P_(1)V_(1))/(4)`
hence, (a) is correct
`Delta=(f)/(2)(P_(2)V_(2)-P_(1)V_(1))`
`=(3)/(2)[(3)/(2)xx2P_(1)V_(1)-P_(1)V_(1)]=3P_(1)V_(1)`
hence, (b) is correct as will Net work done on pistion by all force is zero , So,
`W_(gas)=(1)/(2)Kx^(2)-P(V_(2)-V_(1))=0`
`W_(gas)=(P_(1)V_(1))/(3)+2P_(1)V_(1)`
`W_(gas)=(7)/(3)P_(1)V_(1)`
So, (c) also turns out be correct `Q=DeltaU+W`
`DeltaU=(3)/(2)(P_(2)V_(2)-P_(1)V_(1))=(3)/(2)((4)/(3)P_(1)3V_(1)-P_(1)V_(1))`
`DeltaU=(9)/(2)P_(1)V_(1)`
`Q=(9)/(2)P_(1)V_(1)+(7)/(3)P_(1)V_(1)`
`Q=(41)/(6)P_(1)V_(1)`
Hence (d) in not correct.