Question

A perfect gas is found to obey the relation $p{V}^{\frac{3}{2}}=$ 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:

• A. 2T
• B. 4T
• C. $\sqrt{2}T$
• D. $2\sqrt{2}T$

Answer 1

The correct option is C $\sqrt{2}T$

Given relation ${}_p{V}^{\frac{3}{2}}=$ constant
Adibatic equation $P{V}^y$ = constant
On comparing we get $\gamma =\frac{3}{2}{V}_1=V_0;V_2=\frac{V_0}{2}$
Applying equation for adiabatic process relate V and T.
Important equation used in Adiabtic process $P{V}^{\gamma }=K=$ Constant
$T{V}^{\gamma -1}=K{P}^{1-\gamma }{T}^{\gamma }=K{T}_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}T(V_0{)}^{\gamma -1}=T_2{\left[\frac{V_0}{2}\right]}^{\gamma -1}{T}_2=T(2{)}^{\frac{3}{2}-1}{T}_2=\sqrt{2}T$