Question

An ideal gas (γ = 1.67) is taken through the process abc shown in the figure. The temperature at point a is 300 K. Calculate (a) the temperatures at b and c (b) the work done in the process (c) the amount of heat supplied in the path ab and in the path bcand (d) the change in the internal energy of the gas in the process.

Answer 1

(a) For line ab, volume is constant.
So, from the ideal gas equation,

`"P"_1/"T"_1 = "P"_2/"T"_2`

`=> 100/300 = 200 /"T"_2`

`"T"_2 = ((200 xx 300 )/100) = 600 "K"`

For line bc, pressure is constant.

`So, "V"_1 /"T"_1 = "V"_2 /"T"_2`

`=> 100/600 = 150 / "T"_2`

`=> "T"_2 ((600 xx 150) /100) = 900 "K"`

(b)
As process ab is isochoric, W_ab =0 .

During process bc,
P = 200 kPa

The volume is changing from 100 to 150 cm³ .
Therefore, work done = 50 × 10⁻⁶ × 200 × 10³ J
                                    = 10 J

(c) For ab (isochoric process), work done = 0.
From the first law,
dQ = dU = nC_vdT

⇒ Heat supplied = nC_vdT
Now,

`"Q"_"ab" = (("P""V")/("R""T")) xx (("R")/(gamma - 1)) xx "d""T"`

`= (200 xx 10 ^ 3 xx 100 xx 10 ^-6 xx 300)/(600 xx 0.67)`

= 14.925   ( ∴ γ = 1.67)

F or bc (isobaric process) : 

Heat supplied in bc = `"n" "C"_"p""d""T"      ( "C"_"p" = (gamma "R")/(gamma - 1))`

`= ("P""V")/("R""T") xx (gamma "R") /(gamma -1 ) xx "d""T"`

`=  (200 xx 10^3 xx 150 xx 10 ^-6 )/(600 xx 0.67) xx 300 `

`= 10 xx 1.67 /0.67 = 16.7/0.67 = 24.925`

(d) dQ = dU + dW
Now,
dU = dQ − dW
     = Heat supplied − Work done
     = (24.925 + 14.925) − 10
     = 39.850 − 10 = 29.850 J.