Question

Two samples of gases 1 and 2 are initially kept in the same state. Sample 1 is expanded through an isothermal process whereas sample 2 through an adiabatic process up to the same final volume. The final temperature in process 1 and 2 are $T_1$ and $T_2$ respectively, then

• A. $T_1>T_2$
• B. $T_1=T_2$
• C. $T_1<T_2$
• D. the relation between $T_1$ and $T_2$ cannot be deduced

Answer 1

The correct option is A $T_1>T_2$

The work done is more in isothermal process than adiabatic process.

Let the equal heat given to both be Q.

Work done by 1 is $W_1$.

Work done by 2 is$W_2$.

where $W_1>W_2$

Let the initial temperature be T,

For 1 : the $T_1$ will be same as T because it is an isothermal

process.

$\therefore △v=0$

For 2 : $P{V}^4=K$

$T{V}^{4-1}=K$

$T_1{V}_1^{4-1}=T_2{V}_2^{4-1}$

$\frac{T_1}{T_2}={\left(\frac{V_2}{V_1}\right)}^{4-1}$

As the gas expands : $V_2>V_1;\frac{V_2}{V_1}>1$

$\frac{T_1}{T_2}>1,$

$\therefore T_1>T_2$

Hence, OPTION : A.