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11.Thermodynamics
medium
The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at $27°C$, if the value of $\gamma = 1.4,$ then the new temperature will be
A
$350 \times {4^{0.4}}K$
B
$300 \times {4^{0.4}}K$
C
$150 \times {4^{0.4}}K$
D
None of these
Solution
(b) For adiabatic change $T{V^{\gamma – 1}}$= constant
==> $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma – 1}}$
==> ${T_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma – 1}} \times {T_1}$
==> ${T_2} = {\left( {\frac{V}{{V/4}}} \right)^{1.4 – 1}} \times 300$ $ = 300 \times {(4)^{0.4}}K$
Standard 11
Physics