A perfect gas is found to obey the relation P V3 / 2= constant during an adiabatic process. If such a gas initially at a temperature T, is compressed to half of its initial volume, then its final temperature will be :

Question

A perfect gas is found to obey the relation P V3 / 2= constant during an adiabatic process. If such a gas initially at a temperature T, is compressed to half of its initial volume, then its final temperature will be : (A)  2T  (B)  4T  (C) (2)1/2T  (D)  2(2)1/2T

Tardigrade Admin · Accepted Answer

Gas obeys the equation P V3 / 2= text constant Comparing with P V&gamma;= constant &gamma;=(3/2) For arliabatic process T1 V1&gamma;-1 =T2 V2&gamma;-1 T(V)3 / 2-1 =T2((V/2))3 / 2-1 T2 =T(2)3 / 2-1 =(2)1 / 2 T

Q. A perfect gas is found to obey the relation $P V^{3/2} =$ constant during an adiabatic process. If such a gas initially at a temperature $T$ , is compressed to half of its initial volume, then its final temperature will be :

$2 T$

$4 T$

$(2)^{1/2} T$

$2 (2)^{1/2} T$

Gas obeys the equation
$P V^{3/2} = constant$
Comparing with $P V^{γ} =$ constant
$γ = \frac{3}{2}$
For arliabatic process
$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
$T (V)^{3/2 - 1} = T_{2} (\frac{V}{2})^{3/2 - 1}$
$T_{2} = T (2)^{3/2 - 1}$
$= (2)^{1/2} T$

Q. A perfect gas is found to obey the relation PV3/2= constant during an adiabatic process. If such a gas initially at a temperature T, is compressed to half of its initial volume, then its final temperature will be :

2T

4T

(2)1/2T

2(2)1/2T