Question

A gas initially $15\text{kPa}$, $1.5\text{L}$ and $280\text{K}$ expands at constant temperature until the pressure drops to $12\text{kPa}$. The gas is then held at the constant pressure of $12\text{kPa}$ and heated until the temperature rises to $350\text{K}$. What is the final volume of the gas?

• A. $3.35\text{L}$
• B. $4.50\text{L}$
• C. $2.35\text{L}$
• D. $1.85\text{L}$

Answer 1

The correct option is C $2.35\text{L}$

Given:
At the constant temperature of $280\text{K}$, using Boyle's law:

${\text{P}}_1{\text{V}}_1={\text{P}}_2{\text{V}}_2$

${\text{V}}_2=\frac{{\text{P}}_1}{{\text{P}}_2}{\text{V}}_1$

$⟹\frac{15\times {10}^3}{12\times {10}^3}\times 1.5=\frac{15}{8}\text{L}$

Now, at the constant pressure of $12\text{kPa}$, using Charles' law:

$\frac{{\text{V}}_1^{\ast }}{{\text{T}}_1}=\frac{{\text{V}}_2^{\ast }}{{\text{T}}_2}$

${\text{V}}_2^{\ast }=\frac{{\text{T}}_2}{{\text{T}}_1}{\text{V}}_1^{\ast }$

$⟹\frac{350}{280}\times \frac{15}{8}=\frac{75}{32}$

${\text{V}}_2^{\ast }=2.34375\text{L}\approx 2.35\text{L}$

$\therefore$ The final volume of the gas is $2.35\text{L}$.