Question

Figure illustrates a cycle conducted with $n$ moles of an ideal gas. In the states $a$ and $b$ the gas temperature are $T_a$ and $T_b$ respectively. Temperature of the gas in the state $\text{c}$ is

• A. $\sqrt{T_a{T}_b}$
• B. $T_a+T_b$
• C. $\frac{T_a-T_b}{T_a+T_b}$
• D. $\frac{(T_a+T_b)}{2}$

Answer 1

The correct option is A $\sqrt{T_a{T}_b}$

For process $a\to b$:
From figure, pressure $\left(P\right)$ is directly proportional to the volume $\left(V\right)$.i.e.,
$P\propto V$

So, we can write, $\frac{P_a}{P_b}=\frac{V_a}{V_b}$

Let, $\frac{P_b}{P_a}=\frac{V_b}{V_a}=x$

$\therefore$ $P_b=x{P}_a$ and $V_b=x{V}_a$


Consider the pressure and volume at point $\text{a}$ to be $P_o$ and $V_o$ respectively.

Applying idea gas equation $PV=nRT$ at points $\text{a}$ and $\text{b}$ and substituting the values, we get

$T_a=\frac{P_o{V}_o}{nR}....\left(1\right)$

$T_b=\frac{\left(x{P}_o\right)\left(x{V}_o\right)}{nR}=\frac{x^2{P}_o{V}_o}{nR}....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get

$x=\sqrt{\frac{T_b}{T_a}}$

Now, temperature at point $\text{c}$,

$T_c=\frac{x{P}_o{V}_o}{nR}$

Substituting the values, $T_c=\left(\sqrt{\frac{T_b}{T_a}}\right)T_a$

$\therefore T_c=\sqrt{T_a{T}_b}$

Hence, option (a) is the correct answer.