Question

An ideal gas has volume ${V_0}$ at ${27^ \circ }C$ . It is heated at constant pressure so that its volume becomes $2{V_0}$ . The final temperature is
A. ${54^{\circ} }C$
B. ${32.6^{\circ} }C$
C. ${327^ {\circ} }C$
D. $150K$

Answer 1

Hint: In this problem, an ideal gas is heated ​at constant pressure, resulting in a change in the volume of gas and we know that the temperature will vary with the volume in this situation. In order to determine the final temperature of the gas, it is therefore advised to use the Charles Law, which works at constant pressure.

Formula used:
The formula used in this problem is: -
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$

Complete answer:
It is given that at ${27^ \circ }C$, the volume is ${V_0}$ i.e., the initial volume ${V_1} = {V_0}$ and initial temperature ${T_1} = {27^ \circ }C = 300K$ $\left( {^ \circ C + 273 = K} \right)$

When the gas is heated at constant pressure, the volume becomes double i.e., ${V_2} = 2{V_0}$ (given)

Now, we know that at constant pressure, the volume of gas is directly proportional to its absolute temperature (according to Charles Law).
i.e., $\dfrac{V}{T} = $constant
or, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$ … (1)

Substitute the values given in the question in eq. (1), we get
$ \Rightarrow \dfrac{{{V_0}}}{{2{V_0}}} = \dfrac{{300}}{{{T_2}}}$
$ \Rightarrow {T_2} = 2 \times 300 = 600K$

Converting this into $^ \circ C$, we get
$ \Rightarrow {T_2} = {327^ \circ }C$

Thus, the final temperature when gas is heated at constant pressure will be ${327^ \circ }C$. Hence, the correct option is (C) ${327^ \circ }C$ .

Note: In the given system, to determine the final temperature of an ideal gas, we have to apply the Charles Law. Since this is a numerically based problem on a thermodynamic system with constant pressure, it is crucial to assess the given conditions, such as the change in volume, in order to provide an accurate solution.