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11.Thermodynamics
medium
At ${27^o}C$ a gas is suddenly compressed such that its pressure becomes $\frac{1}{8}th$ of original pressure. Temperature of the gas will be $(\gamma = 5/3)$
A
$420K$
B
${327^o}C$
C
$300K$
D
$ - {142^o}C$
Solution
(d) ${T^\gamma }{P^{1 – \gamma }} = $constant ==> $T \propto {P^{\frac{{\gamma – 1}}{\gamma }}}$
==> $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma – 1}}{\gamma }}} = {\left( {\frac{1}{8}} \right)^{\frac{{5/3 – 1}}{{5/3}}}}$
${T_2} = 300 \times {\left( {\frac{1}{8}} \right)^{0.4}} = 131K = – 142^\circ C$
Standard 11
Physics
Similar Questions
medium
In Column$-I$ process and in Column$-II$ first law of thermodynamics are given. Match them appropriately :
Column$-I$ | Column$-II$ |
$(a)$ Adiabatic | $(i)$ $\Delta Q = \Delta U$ |
$(b)$ Isothermal | $(ii)$ $\Delta Q = \Delta W$ |
$(iii)$ $\Delta U = -\Delta W$ |
easy