A sample of gas X occupies 10 $m^{3}$ $m^3$ at the pressure of 120 kPa. If the volume of the sample is 3 $m^{3}$ $m^3$ , what is the new pressure of the gas?

Question

8005 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-21T03:16:43

The new pressure is 400 kPa

To solve this problem use Boyle's law.

As the volume decreases the pressure increases. This means that the new pressure will be greater than the original pressure.

$V_{1} \times P_{1} = V_{2} \times P_{2}$ $V_1 xx P_1 = V_2 xx P_2$

$V_{1} = 10 m^{3}$ $V_1 = 10m^3$
$P_{1} = 120 k P a$ $P_1 = 120 kPa$
$V_{2} = 3 m^{3}$ $V_2 = 3m^3$
$P_{2} = ?$ $P_2 = ?$

Putting these values into the equation gives.

$10 m^{3} \times 120 k P a = 3 m^{3} \times ?$ $10m^3 xx 120 kPa = 3m^3 xx ?$

Dividing both sides by $3 m^{3}$ $3m^3$ gives

$\frac{10 m^{3} \times 120 k P a}{3 m^{3}} = \frac{3 m^{3} \times ?}{3 m^{3}}$ $(10m^3 xx 120kPa)/(3m^3) = (3m^3 xx ?)/(3m^3)$ This gives

400 kPa = ?

The unknown is 400 kPa.

A sample of gas X occupies 10m^3m3m^3 at the pressure of 120 kPa. If the volume of the sample is 3m^3m3m^3, what is the new pressure of the gas?