Question

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

at STP?
O
49123.7 mL
3.02 mL
2415.0 mL
612.8 mL

Answer 1

The new volume is 2415 mL

Step-by-step 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:

`(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:

`(P)/(T) =k`

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

`(P*V)/(T) =k`

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

`(P1*V1)/(T1) =(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:

`(0.9 atm*4.6L)/(468K) =(1 atm*V2)/(273K)`

Solving:

`V2=(0.9 atm*4.6L)/(468K)*(273K)/(1 atm)`

V2= 2.415 L =2,415 mL

The new volume is 2415 mL