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A mas occupies $11.2 \mathrm{dm}^3$ at $105 \mathrm{kPa}$. What is its volume if pressure is increased to $210 \mathrm{kPa}$ ?
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Verified Answer
The correct answer is:
$5.6 \mathrm{dm}^3$
$$
\mathrm{P}_1=105 \mathrm{kPa}, \mathrm{V}_1=11.2 \mathrm{dm}^3
$$
$$
\mathrm{P}_2=210 \mathrm{kPa}, \mathrm{V}_2=?
$$
According to Boyle's law, $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{P}_2$
$$
\therefore \mathrm{V}_2 \frac{\mathrm{P}_1 \mathrm{~V}_2}{\mathrm{P}_2}=\frac{105 \mathrm{kPa} \times 11.2 \mathrm{dm}^3}{210 \mathrm{kPa}}=5.6 \mathrm{dm}^3
$$
\mathrm{P}_1=105 \mathrm{kPa}, \mathrm{V}_1=11.2 \mathrm{dm}^3
$$
$$
\mathrm{P}_2=210 \mathrm{kPa}, \mathrm{V}_2=?
$$
According to Boyle's law, $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{P}_2$
$$
\therefore \mathrm{V}_2 \frac{\mathrm{P}_1 \mathrm{~V}_2}{\mathrm{P}_2}=\frac{105 \mathrm{kPa} \times 11.2 \mathrm{dm}^3}{210 \mathrm{kPa}}=5.6 \mathrm{dm}^3
$$
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