Half mole of an ideal gas (γ = 5/3) is taken through the cycle abcda, as shown in the figure. Take `"R" = 25/3"J""K"^-1 "mol"^-1 `. (a) Find the temperature of the gas in the states a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c) Find the amount of heat liberated in the processes cd and da.

#### Solution

Given:

Number of moles of the gas,

n = 0.5 mol

`"R" = 25/3 `J/mol -K

`gamma =5/3`

(a) Temperature at a = T_{a}

P_{a}V_{a} =nRT_{a}

`=> "T""a" = ("P"_"a""V"_"a")/("n""R") = (100 xx 10^3 xx 5000 xx 10^-6)/(0.5 xx 25/3) = 120 "K"`

Similarly, temperature at b,

`"T"_"b" = ("P"_b"V"_"b")/("n""R")`

`"T"_"b" =( 100 xx 10^3 xx 10000 xx 10^-6)/(0.5 xx 25/3)`

T_{b} =240 K

Similarly, temperature at c is 480 K and at d is 240 K.

(b) For process ab,

dQ = nc_{p}dT

[Since ab is isobaric]

`"d""Q" = 1/2 xx ("R"gamma)/(gamma-1) ("T"_"b" - "T"_"a")`

`"d""Q" =1/2 xx ((25 xx 5)/(3 xx 3))/(5/3 -1 ) xx (240 -120 )`

`"d""Q" = 1/2 xx 125 /9 xx 3/2 xx (120)`

dQ = 1250 J

For line bc, volume is constant. So, it is an isochoric process.

dQ = dU + dW

[dW = 0, isochoric process]

dQ = dU = nC_{v}dT

dQ = nC_{v} (T_{c} - T_{b})

`"d""Q" = 1/2 xx ((25/3))/[[(5/3)-1]] xx (240)`

`"d""Q" = 1/2 xx 25/3 xx 3/2 xx 240 =1500 "J"`

(c) Heat liberated in cd (isobaric process),

dQ = − nC_{p}dT

`"d""Q" = -1/2 xx( gamma "R")/(gamma-1) xx ("T"_d -"T"_"c")`

`"d""Q" = -1/2 xx 125/9 xx 3/2 xx ( 240 -480)`

`"d""Q" = -1/2 xx 125/6 xx 240 = 2500 "J"`

Heat liberated in da (isochoric process),

⇒ dQ = dU

Q= −nC_{v}dT

`"d""Q" = -1/2 xx "R" /(gamma -1 )("T"_"a"-"T"_"d")`

`"d""Q" = -1/2 xx 25/2 xx (120-240)`

`"d""Q" = 25/4 xx 120 =750 "J"`