Question

V2 = (P1 × V1 × T2) (P2 × T1) A gas with a beginning pressure of 2 atm at a temperature of 300 K has a volume of 20 ml. What will be the new volume if the pressure increases to 4 atm and the temperature is lowered to 200 K?

Answer 1

6.6ml will be the new volume if the pressure increases to 4 atm and the temperature are lowered to 200 K.

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given data:

$V_1=20 ml$

$T_1=3$

$P_1=2 atm$

$T_2=200 K$

$P_2=4 atm$

$V_2=?$

Using equation:

$V_2 =$ $\frac{P_1 X V_1 XT_2}{P_2 X T_1}$

$V_2 =$$\frac{2 atm X 20 ml X200 K}{4 atm X 300 K}$

$V_2 = 6.6 ml$

Hence, 6.6ml will be the new volume if the pressure increases to 4 atm and the temperature are lowered to 200 K.

Learn more about the ideal gas equation here:

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