A gas occupies 2.0 m3 at 100.0k and exerts a pressure of 100.0kPa. What volume will the gas occupy if the temperature is increased to 400.0 K and the pressure is increased to 200.0kPa

Answer :

PBCHEM
According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure,  T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have
[tex] \frac{P1V1}{P2V2} = \frac{T1}{T2} [/tex]

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa

∴ on rearranging above eq., we get V2 = [tex] \frac{P1V1T2}{T1} = \frac{100 X 2 X 400}{200X100} [/tex]
∴ V2 = 4 m3 

The ideal equation relates the temperature with the pressure and the volume of the gas. When the temperature is increased then the volume will be 4 cubic meters.

What is an ideal gas equation?

An ideal gas equation depicts the relation between the temperature to that of the volume and the pressure of the gas.

The formula is given as,

[tex]\rm \dfrac{P_{1}V_{1}}{P_{2}V_{2}} = \rm \dfrac{T_{1}}{T_{2}}[/tex]

Given,

Initial pressure = 100 kPa

Initial volume = 2 cubic meter

Initial temperature = 100 K

Final pressure = 200 kPa

Final volume = ?

Final temperature = 400 K

The final volume is calculated as:

[tex]\begin{aligned} \rm V_{2} &= \rm \dfrac{P_{1}V_{1}T_{2}}{T_{1}}\\\\&= \dfrac{100\times 2 \times 400}{200 \times 100}\\\\&= 4 \;\rm m^{3}\end{aligned}[/tex]

Therefore, 4 cubic meters is the volume of the final gas.

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