Question

If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml

Answer 1

Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

$\frac{V}{T} =k$

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

$\frac{P}{T} =k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T} =k$

Having two different states, an initial state and an final state, it is true:

$\frac{P1*V1}{T1} =\frac{P2*V2}{T2}$

In this case:

• P1= 0.9 atm
• V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
• T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

• P2= 1 atm
• V2= ?
• T2= 0°C= 273 °K

Replacing:

$\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}$

Solving:

$V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}$

V2= 2.415 L =2,415 mL

The new volume is 2415 mL