1: For a given amount, type, and temperature of gas, when its pressure is increased from 200 kPa to 500 kPa, what will its volume be if its initial volume is 3 cubic meters?

Question

Mateo Aguilar · Accepted Answer

#"1.2 m"^3#

#P*V=n*R*T " " #(Ideal gas law)

The product of pressure and volume is constant at any given time, assuming that the temperature and gas amount are fixed. You can use this to find the straightforward equation:

For #P_1# and #V_1# as the initial pressure and volume and #P_2# and #V_2# as the pressure and volume at any given moment.

#P_1 xx V_1 = P_2 xx V_2#

Writing down the values gives us:

#"3 m"^3 xx "200 kPa" = V_2 xx "500 kPa"#

Thus

#"600 kPa m"^3 = V_2 xx "500 kPa"#

#V_2 = "600 kPa m"^3/"500 kPa" = 6/5 quad "m"^3 = "1.2 m"^3#

The volume is still expressed in cubic meters because we only worked with numbers.

Noah Aldrich · Answer

undefined Using the ideal gas law equation $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature:

Given:
Initial pressure ($P_1$) = 200 kPa
Final pressure ($P_2$) = 500 kPa
Initial volume ($V_1$) = 3 cubic meters

Using the ideal gas law equation and rearranging for the final volume ($V_2$):

$$V_2 = \frac{{P_1 \times V_1}}{{P_2}}$$

$$V_2 = \frac{{200 \times 3}}{{500}} = 1.2 \text{ cubic meters}$$

So, the final volume of the gas will be 1.2 cubic meters.

Peyton Adams · Answer

undefined Use Boyle's Law to find the final volume. Boyle's Law states that for a given amount and temperature of gas, the pressure multiplied by the volume is constant. Therefore, $P_1V_1 = P_2V_2$, where $P_1$ and $V_1$ are the initial pressure and volume, and $P_2$ and $V_2$ are the final pressure and volume, respectively. Rearrange the equation to solve for $V_2$. Then substitute the given values to find the final volume $V_2$.