A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If l1 and l2 are the lengths of the gas column, before and after the expansion respectively, then the value of T_1 \over T_2 will be :

Heat and Thermodynamics PYQ — JEE Main 2021 · 25 Jul · Shift 1 · Q53 with Solution

View written solutionFree

Correct answer: B

The gas expands adiabatically, so for an ideal gas:

$PV^\gamma=\text{constant}$

Also,

$TV^{\gamma-1}=\text{constant}$

For a monoatomic ideal gas,

$\gamma=\frac{C_P}{C_V}=\frac{5}{3}$

Hence,

$\gamma-1=\frac{5}{3}-1=\frac{2}{3}$

So the adiabatic relation becomes:

$TV^{2/3}=\text{constant}$

The gas is enclosed in a cylinder of constant cross-sectional area $A$ .

Therefore,

$V=A l$

where $l$ is the length of the gas column.

Thus,

$V_1=A l_1, \qquad V_2=A l_2$

Using

$T_1 V_1^{2/3}=T_2 V_2^{2/3}$

we get

$T_1 (A l_1)^{2/3}=T_2 (A l_2)^{2/3}$

The factor $A^{2/3}$ cancels:

$T_1 l_1^{2/3}=T_2 l_2^{2/3}$

Therefore,

$\frac{T_1}{T_2}=\left(\frac{l_2}{l_1}\right)^{2/3}$

This corresponds to:

$\boxed{\left(\frac{l_2}{l_1}\right)^{2/3}}$

So the correct option is B.

Stored correct answer: B

Our derived answer: B

They match.