#### A bulb of three litre capacity filled with air is heated from $\dpi{100} \small 27^{\circ}C$ to $\dpi{100} \small t^{\circ}C$. The air thus expelled measured 1.45L at $\dpi{100} \small 17^{\circ}C$. Considering the pressure to be 1 atm throughout the experiment and ignoring the expansion of bulb, calculate temprature (in oC).Option: 1 327Option: 2 320Option: 3 350Option: 4 335

We have:
Initial volume of bulb = 3L
Initial temperature = 273 + 27 = 300K
Final temperature = 273 + t
Thus, according to charles’ law, we have:

$\mathrm{\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}}}$

Now at $17^{\circ}C$, the volume of gas is 1.45L, thus, we have:

$\mathrm{\frac{1.45}{290} =\frac{V}{t\: +\:273}}$...............................(i)

Now the gas is heated from $27^{\circ}C$ to $t^{\circ}C$, thus volume can be given as follows:

$\mathrm{\frac{3}{300}\: =\: \frac{3\, +\, V}{273\, +\, V}\: =\: \frac{V}{273\, +\, t}\: +\: \frac{3}{273\, +\, t}}$

Using equation (i) we have:

$\mathrm{\frac{3}{300}\: =\: \frac{3\, +\, V}{273\, +\, V}\: =\: \frac{1.45}{290}\: +\: \frac{3}{273\, +\, t}}$

Thus, $t=327^{\circ}C$