0.2moles of an ideal gas, is taken around the cycle abc as shown in the figure. The path b-c is adiabatic process, a-b is isovolumic process and c-a is isobaric process. The temperature at a and b are `T_(A)=300K` and `T_(b)=500K` and pressure at a is 1 atmospere. find the volume at c.
(Given, `gamma=(C_(p))/(C_(v))=(5)/(3),R=8.205xx10^(-2)L`/atm/mol-K)

Options

A. 6.9L
B. 6.68L
C. 5.52L
D. 5.82L

Answer

b

Explanation

From a-b, volume is constnat
`(p_(s))/(T_(a))=(P_(b))/(T_(b))`
`P_(b)=(500)/(300)xx1=(6)/(3)` atm
For b-c, adiabatic process
`(T_(c))/(T_(b))=((P_(c))/(P_(b)))^((gamma-1)/(gamma))=[(1)/(5//3)]^((5//3-1)/(5//3))=((3)/(5))^(2//5)`
Also, `p_(b)V_(b)^(gamma)=p_(c)V_(c)^(gamma)`
`V_(c)=(V_(b))((p_(b))/(p_(c)))^(1//gamma)`
`=(4.928)((5//3)/(1))^(3//5)`=6.68litre