Question

A monatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The volume ratios are $\left(V_B/V_A\right)=2$ and $\left(V_D/V_A\right)=4$. If the temperature $T_A$ at A is ${27}^o$C, calculate the temperature in K of the gas at point B.

Answer 1

Given: $\frac{V_B}{V_A}=2$

$T_A={27}^{0}C$

Solution: As we know AB is a straight line in a V-T graph.

$\therefore \frac{V}{T}=Constant$ (iso-choric process)

$\frac{V_A}{T_A}=\frac{V_B}{T_B}$

$T_B=\frac{V_B}{V_A}\times T_A$

$T_B=2\times 300$

$T_B=600K$

Hence, the correct answer is 600K.