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An ideal gas after going through a serie...

An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state again (cyclic process). The amounts of heat (Q) and work (W) involved in the states are,
`Q_(1) = 6000J, Q_(2)= -5500 J,`
`Q_(3)=-3000 J, Q_(4)=3500 J`
`W_(1)=2500J, W_(2)=-1000 J,`
`W_(3)=-1200 J, W_(4)=xxJ`
The ratio of net work done by the gas to the total heat absorbed by the gas in η. The value of x and `eta` are nearly :-

A

500,7.5 %

B

700,10.5 %

C

1000,21 %

D

1500,15%

Text Solution

Verified by Experts

The correct Answer is:
B
From first law of thermodynamics
`Q=Delta U+W`
or `Delta U=Q-W`
Therefore , `Delta U_(1)=Q_(1)-W_(1)=6000-2500=3500 J`
`Delta U_(2)=Q_(2)-W_(2)=-5500+1000=-4500 J`
`Delta U_(3)=Q_(3)-W_(3)=-3000+1200=-1800 J`
`Delta U_(4)=Q_(4)-W_(4)=3500 -x=0`
For cyclic process `Delta U=0`
3500-4500-1800+3500-x=0
or x=700 J
Efficiency. `eta=("output")/("input")xx100`
`=(W_(1)+W_(2)+W_(3)+W_(4))/(Q_(1)+Q_(4))xx 100`
`=(2500-1000-1200+700)/(6000+3500)xx 100`
`=(1000)/(9500)xx 100`
`eta=10.5%`
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