Question

A bulb of tree litre capacity filled with air is headed from ${27}^{o}C$ to $t^{o}C$. The air thus expelled measured $1.45$ litre at ${17}^{o}C$ considering the pressure to be $1$ atm, throughout the experiment and ignoring the expansion of bulb, calculate $t$.

Answer 1

$\frac{P_{300}{V}_{300}}{n_{300}300}=\frac{P_t{V}_t}{n_{t}t}$
$P_{300}=P_t=1atm$
$V_{300}=3L$
$n_{300}=n_t$
$So,\frac{3}{300}=\frac{3+x}{t}$
$Rearrangethem$
$t=\frac{300\times (3+x)}{3}------\left(1\right)$
$Given:$
$\frac{V_{290}}{T_{290}}=\frac{V_t}{t}$
$\frac{1.45}{290}=\frac{x}{t}$
$x=\frac{1.45\times t}{290}----------\left(2\right)$
Put the value of x from equation (2) to euation(1) and solve for t
$t=\frac{300\times (3+\frac{1.45t}{290})}{3}$
$=\frac{300\times (3+(5\times 10^{-3}\times t)}{3}$
$=\frac{900+1.5t}{3}=\frac{900}{3}+\frac{1.5t}{3}$
$t-\frac{1.5t}{3}=\frac{900}{3}\frac{3t-1.5t}{3}=300$
$\frac{1.5}{3}\times t=300$
$0.5\times t=300$
$t=\frac{300}{0.5}=600K$
$\frac{P_{300}{V}_{300}}{n_{300}300}=\frac{P_t{V}_t}{n_{t}t}$
$P_{300}=P_t=1atm$