A monatomic 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 length of the gas column before and after expansion respectively, then (T1/T2) is given by

Question

A monatomic 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 length of the gas column before and after expansion respectively, then (T1/T2) is given by (A) ((L1/L2))2 / 3 (B) (L1/L2) (C) (L2/L1) (D) ((L2/L1))2 / 3

Tardigrade Admin · Accepted Answer

Here T V&gamma;-1= constant As &gamma;=(5/3), hence T V2 / 3= constant Now T1 L12 / 3=T2 L22 / 3(∵ V propto L); Hence, (T1/T2)=((L2/L1))2 / 3

$(\frac{L _{1}}{L _{2}})^{2/3}$

$\frac{L _{1}}{L _{2}}$

$\frac{L _{2}}{L _{1}}$

$(\frac{L _{2}}{L _{1}})^{2/3}$

Here $T V^{γ - 1} =$ constant
As $γ = \frac{5}{3}$ , hence $T V^{2/3} =$ constant
Now $T_{1} L_{1}^{2/3} = T_{2} L_{2}^{2/3} (∵ V \propto L)$ ;
Hence, $\frac{T _{1}}{T _{2}} = (\frac{L _{2}}{L _{1}})^{2/3}$

(L2​L1​​)2/3

L2​L1​​

L1​L2​​

(L1​L2​​)2/3

Here TVγ−1= constant As γ=35​, hence TV2/3= constant Now T1​L12/3​=T2​L22/3​(∵V∝L); Hence, T2​T1​​=(L1​L2​​)2/3

$(\frac{L _{1}}{L _{2}})^{2/3}$

$\frac{L _{1}}{L _{2}}$

$\frac{L _{2}}{L _{1}}$

$(\frac{L _{2}}{L _{1}})^{2/3}$

Here $T V^{γ - 1} =$ constant
As $γ = \frac{5}{3}$ , hence $T V^{2/3} =$ constant
Now $T_{1} L_{1}^{2/3} = T_{2} L_{2}^{2/3} (∵ V \propto L)$ ;
Hence, $\frac{T _{1}}{T _{2}} = (\frac{L _{2}}{L _{1}})^{2/3}$