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11.Thermodynamics
medium
A tyre filled with air $({27^o}C,$ and $2$ atm) bursts, then what is temperature of air ....... $^oC$ $(\gamma = 1.5)$
A
$ - 33$
B
$0$
C
$27$
D
$240$
Solution
(a) $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma – 1}}{\gamma }}}$==> $\frac{{{T_2}}}{{(273 + 27)}} = {\left( {\frac{1}{2}} \right)^{\frac{{1.5 – 1}}{{1.5}}}} = $${\left( {\frac{1}{2}} \right)^{\frac{1}{3}}} = \frac{1}{{2.5}}$
==> ${T_2} = \frac{{{T_1}}}{{1.25}} = \frac{{(273 + 27)}}{{1.25}} = 238\,K = – \,34.8^\circ C$
Standard 11
Physics
Similar Questions
Match the thermodynamic processes taking place in a system with the correct conditions. In the table: $\Delta Q$ is the heat supplied, $\Delta W$ is the work done and $\Delta U$ is change in internal energy of the system
| Process | Condition |
| $(I)$ Adiabatic | $(A)\; \Delta W =0$ |
| $(II)$ Isothermal | $(B)\; \Delta Q=0$ |
| $(III)$ Isochoric | $(C)\; \Delta U \neq 0, \Delta W \neq 0 \Delta Q \neq 0$ |
| $(IV)$ Isobaric | $(D)\; \Delta U =0$ |
