Question

In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at $ T_1 = 300\,K$ and $ T_2 = 200\,K$. The value of $ V_A = 2$ unit, $ V_B = 8$ unit, $V_C = 16$ unit. Find the value of $ V_D$.

• A. 4 unit
• B. < 4 unit
• C. > 5 unit
• D. 5 unit

Answer 1

Answer: (A)

$T_1 = 300\,K$
$T_2 = 200\,k$
$V_A = 2, V_B = 8, V_C = 16$
$T_BV_B^{r-1} = T_CV_C^{r-1}$
$T_A V_A^{r-1} = T_DV_D^{r-1}$ (adiabatic process)
and $T_1V_B^{r - 1} = T_2 V_C^{r-1}$
$T_1V_A^{r-1}T_2V_D^{r-1}$
$\Rightarrow (300)8^{r-1} = 200\cdot 16^{r-1}$
$ 300(2)^{r-1} = 200 V_d^{r-1}$
$\Rightarrow \frac{300}{200} = 2^{r-1}$
$\Rightarrow \frac{300}{200} = (\frac{V_D}{2})^{r-1}$
$\Rightarrow 2^{r-1} = 1.5 $
$\Rightarrow 1.5 = (\frac{V_D}{2})^{r-1}$
$\Rightarrow 2^{r-1} = (\frac{V_D}{2})^{r-1}$
$\Rightarrow 2 = \frac{V_D}{2}$
$\Rightarrow V_D = 4\,$ units