Question

Air is filled in a motor tube at 27 ${}^\circ C$ and a pressure of 2 atmospheres. The tube suddenly bursts, then the temperature of the air is?
(A) 642 K
(B) 563 K
(C) 300 K
(D) 246 K

Answer 1

We are given that air is filled in a tyre tube at a given temperature and pressure. Then suddenly it bursts, since this process is very fast, so it will fall in the category of adiabatic process. We can use the equation of adiabatic process to solve this problem.

Complete step by step answer:
Given temperature is in ${}^\circ C$first of all we have to convert it into Kelvin. Always the temperatures are in Kelvin but if we have to make a difference then there is no need to convert the units.
27 ${}^\circ C$= 27+273
=300K
Also given pressure is 2 atm. We use here equation $T{{P}^{\gamma }}=K$, where K is a constant
Also, after bursting, the temperature we do not know but the pressure equals atmospheric pressure.
$\dfrac{{{T}_{2}}}{{{T}_{1}}}={{[\dfrac{{{P}_{2}}}{{{P}_{1}}}]}^{\dfrac{\gamma -1}{\gamma }}}$
We know $\gamma$ for air is 1.4

& \dfrac{{{T}_{2}}}{300}={{[\dfrac{1}{2}]}^{\dfrac{1.4-1}{1.4}}} \\\ & {{T}_{2}}=300\times {{0.5}^{0.28}} \\\ & =246K \\\ \end{aligned}$$ So, the value of temperature comes out to be 246 K **Additional Information:** $$\gamma $$= 1.4 for air is taken from the standard values table. It is a diatomic gas. **Note:** After the burst, the pressure becomes equal to atmospheric pressure. The adiabatic process does not involve any transfer of heat to or inside the system. $$\gamma $$is the ratio of specific heat at constant pressure to specific heat at constant volume. It is also given by a relation called Mayer’s formula.