Question
A perfect gas at $${27^ \circ }C$$ is heated at constant pressure so as to double its volume. The final temperature of the gas will be, close to
A.
$${327^ \circ }C$$
B.
$${200^ \circ }C$$
C.
$${54^ \circ }C$$
D.
$${300^ \circ }C$$
Answer :
$${327^ \circ }C$$
Solution :
Given, $${V_1} = V,{V_2} = 2V$$
$${T_1} = {27^ \circ } + 273 = 300\,K\,{\text{so,}}\,{T_2} = ?$$
From charle’s law
$$\eqalign{
& \frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\,\,{\text{or,}}\,\,\frac{V}{{300}} = \frac{{2V}}{{{T_2}}}\,\,\left( {\because {\text{Pressure is constant}}} \right) \cr
& \therefore {T_2} = 600\,K = 600 - 273 = {327^ \circ }C \cr} $$