Question

One mole of an ideal gas undergoes an isochoric
change from A to B and then an isobaric change
from B to C. The temperature at A and also at C
is 27°C. Find the total amount of heat absorbed
by the gas in the entire process.
(1) 150 R (2) 400 R (3) 250 R (4) 300 R​

Answer 1

Given :

One mole of an ideal gas undergoes an isochoric  change from A to B and then an isobaric change  from B to C

The temperature at A = 27° c

The temperature at B = 27° c

To Find :

Total amount of heat absorbed

Solution :

Let The pressure at point A = P

And The volume at point A = V

Using Ideal gas equation at A

i.e  P V = n R T                where number of mole =  n = 1

Or, , P V = R T                      .............(1)

∵      From point A to point b , there is a isochoric process

So, At isochoric process , volume = constant

And Temperature at point B = $T_B$  = $\dfrac{T}{n}$              (given)

∴     At constant volume

      $\dfrac{P_A}{T_A}$  = $\dfrac{P_B}{T_B}$

Or, $\dfrac{P_A}{T}$  =  $\dfrac{P_B}{\dfrac{T}{n} }$

Or, $P_A$  = n $P_B$

And Pressure at point B = $P_B$  = $\dfrac{P_A}{n}$         ............2

Again

From point B to Point C , There is Isobaric process

So, At Isobaric process , pressure = constant

Thus    pressure at point C = pressure at point B

I.e         $P_C$  =  $P_B$  = $\dfrac{P_A}{n}$  

And        $\dfrac{V_B}{T_B}$  =  $\dfrac{V_C}{T_C}$

∵       $T_B$  = $\dfrac{T}{n}$

So,            $\dfrac{V_B}{\dfrac{T}{n} }$  =  $\dfrac{V_C}{T_C}$

∴          $V_C$  = n $V_B$              

Now,

As the initial and final temperature of the gas is same, thus the change in internal energy after both the processes is zero

i.e  Δ$U_A_C$  = 0                .............3

So,  Total work done in going from A to C

i.e     $W_A_C$ = $W_A_B$  + $W_B_C$

               = 0 + $\dfrac{P}{n}$  ( $V_C$  - $V_B$ )

               =  0 + $\dfrac{P}{n}$  ( n V  - V )

​         =  0 + P V ( 1 - $\dfrac{1}{n}$ )

i.e   $W_A_C$ =  P V ( 1 - $\dfrac{1}{n}$ )                   ...........4

From The law of Thermodynamics

    Heat absorbed = change in internal energy + change in work done

i.e    Δ$Q_A_C$  =  Δ$U_A_C$  + Δ$W_A_C$

Or,               = 0 + P V ( 1 - $\dfrac{1}{n}$ )                    ( from eq 3 and eq 4 )

or,       Δ$Q_A_C$   = P V ( 1 - $\dfrac{1}{n}$ )    

Or,     Δ$Q_A_C$   = RT ( 1 - $\dfrac{1}{n}$ )                  ( from eq 1)

Or,            Δ$Q_A_C$   = RT  

For   T = 27°

i.e     T = 273°  +  27°  = 300 kelvin

∴        Δ$Q_A_C$   = 300 R

Hence, The total amount of heat absorbed  by the gas in the entire process is 300 R   Answer

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