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 bc and (d) the change in the internal energy of the gas in the process.
Figure

Answer 1

(a) the temperature at 'b' and 'c' is 900 K

(b) The work done in a process 10 J

(C) The amount of heat supplied is 24.925 J

(d) The change in internal energy of the gas in the process is 29.850 J

Explanation:

Given data

Temperature of the gas = 300 K

γ=1.67

(a) The temperature at 'b' and 'c'

For line ab, volume is constant.

The ideal equation for gas,

$P_1 T_1=P_2 T_2$

$\frac{100}{300} = \frac{200}{T_2}$

$T_{2}=\frac{200 \times 300}{100}=600 K$

For line bc, pressure is constant.

$\frac{V_{1}}{T_{2}}=\frac{V_{2}}{T_{2}}$  

$\frac{100}{600}=\frac{150}{T_{2}}$

$T_{2}=\frac{150 \times 600}{100}$

$T_{2}=900 K$

(b) The work done in the process

Isochoric as method ab,   $W_(ab)=0$

In bc process,

P = 200 K Pa  

The volume varies from 100 to 150 cm³  .

Therefore,  

$\text { work done }=50 \times 10^{-6} \times 200 \times 10^{3} J$

Workdone = 10 J

(c) The amount of heat supplied in the path ab and in the path bc

In case of ab (isochoric process),   work done = 0.  

From the first law,

$d Q=d U=n C_{v} d T$

$\text {Heat supplied}=n C_{v} d T$

$Q_{a b}=\left(\frac{P V}{R T}\right) \times \frac{R}{\gamma-1} \times d T$

      $=\frac{200 \times 10^{3} \times 100 \times 10^{-6} \times 300}{600 \times 0.67}$

$Q_{a b}=14.925$      where   (∴γ=1.67)

For bc (isobaric process):

$\text {Heat supplied in } b c=n C_{p} d T \quad\left(C_{p}=\frac{\gamma R}{\gamma-1}\right)$

                              $=\frac{P V}{R T} \times \frac{\gamma R}{\gamma-1} \times d T$

                              $=\frac{200 \times 10^{3} \times 150 \times 10^{-6} \times 300}{600 \times 0.67}$

                              $=10 \times \frac{1.67}{0.67}=\frac{16.7}{0.67}=24.925$

$\text {Heat Supplied in } b c=24.925$ J

(d) The change in internal energy of the gas

dQ = dU + dW

Now,  

dU = dQ – dW  

    = Heat supplied - Work done  

    = (24.925 + 14.925) – 10  

    = 39.850 - 10

dU  = 29.850 J.

Therefore for an ideal gas with temperature at a point is 300 K (a) the temperature at 'b' and 'c' is 900 K (b) the work done in the Isochoric process  10 J (c) The amount of heat supplied is 24.925 J (d) The change in internal energy of the gas in this process is 29.850 J