Question

the image above shows a chamber with a fixed volume filled with gas at a pressure of 1560 mmHg and a temperature of 445.0 K. If the temperature drops to 312.0K what is the new pressure of the gas in the chamber

Answer 1

Answer:

The new pressure of the gas in the chamber is 1,093.75 mmHg

Explanation:

The Gay-Lussac Law is a gas law that relates pressure and temperature to constant volume. This law says that the pressure of the gas is directly proportional to its temperature.

That is, if the temperature increases, the pressure increases, while if the temperature decreases, the pressure decreases. So the Gay-Lussac law can be expressed mathematically as follows:

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

Having an initial and an end state of a gas, the following expression can be used:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

• P1= 1560 mmHg
• T1= 445 K
• P2=?
• T2= 312 K

Replacing:

$\frac{1560 mmHg}{445 K} =\frac{P2}{312 K}$

Solving:

$P2=\frac{1560 mmHg}{445 K} *312K$

P2=1,093.75 mmHg

The new pressure of the gas in the chamber is 1,093.75 mmHg