Question

In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at $T_1=300K$ and $T_2=200K$. 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)

$T_1=300K$
$T_2=200k$
$V_A=2,V_B=8,V_C=16$
$T_B{V}_B^{r-1}=T_C{V}_C^{r-1}$
$T_A{V}_A^{r-1}=T_D{V}_D^{r-1}$ (adiabatic process)
and $T_1{V}_B^{r-1}=T_2{V}_C^{r-1}$
$T_1{V}_A^{r-1}{T}_2{V}_D^{r-1}$
$\Rightarrow (300)8^{r-1}=200\cdot 16^{r-1}$
$300(2{)}^{r-1}=200V_d^{r-1}$
$\Rightarrow \frac{300}{200}=2^{r-1}$
$\Rightarrow \frac{300}{200}=(\frac{V_D}{2}{)}^{r-1}$
$\Rightarrow 2^{r-1}=1.5$
$\Rightarrow 1.5=(\frac{V_D}{2}{)}^{r-1}$
$\Rightarrow 2^{r-1}=(\frac{V_D}{2}{)}^{r-1}$
$\Rightarrow 2=\frac{V_D}{2}$
$\Rightarrow V_D=4$ units