Question

If I have 3.0 L of a gas at a pressure of 500kPa and a temperature of 50C and I decrease the pressure to 240 kPa and the temperature to 103K what will be the new volume

Answer 1

Answer:

The new volume is 1.993 L

Explanation:

The gas laws are a set of chemical and physical laws that allow determining the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature and moles.

Boyle's law says that the volume occupied by a certain gaseous mass at constant temperature is inversely proportional to the pressure. This law is expressed mathematically as:

P*V=k

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature. This law indicates mathematically that the quotient between pressure and temperature is constant:

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

Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

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

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

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

Analyzing an initial state 1 and a final state 2, it is satisfied:

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

In this case:

• P1= 500 kPa
• V1= 3 L
• T1= 50 C= 323 K (being 0 C= 273 K)
• P2= 240 kPa
• V2= ?
• T2= 103 K

Replacing:

$\frac{500 kPa*3 L}{323 K}=\frac{240 kPa*V2}{103 K}$

Solving:

$V2=\frac{103 K}{240 kPa} *\frac{500 kPa*3 L}{323 K}$

V2=1.993 L

The new volume is 1.993 L