Question

Question 6

Answer 1

Considering the combined law equation, the new temperature is -244.56 °C or 28.44 K.

Boyle's law

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant: if the pressure increases, the volume decreases while if the pressure decreases, the volume increases.

Mathematically, this law states that the multiplication of pressure by volume is constant:

P×V=k

Charles's law

Charles's law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles' law states that the ratio of volume to temperature is constant:

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

Gay-Lussac's law

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature: increasing the temperature will increase the pressure, while decreasing the temperature will decrease the pressure.

Mathematically, Gay-Lussac's law states that the ratio of pressure to temperature is constant:

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

Combined law equation

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

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

Considering an initial state 1 and a final state 2, it is fulfilled:

$\frac{P1xV1}{T1} =\frac{P2xV2}{T2}$

New temperature

In this case, you know:

• P1= 1 atm= 760 mmHg
• V1= 15 L
• T1= -30 °C= 243 K (being 0 °C= 273 K)
• P2= 58 mmHg
• V2= 23 L
• T2= ?

Replacing in the combined law equation:

$\frac{760 mmHgx15 L}{243 K} =\frac{58 mmHgx23 L}{T2}$

Solving:

$T2x\frac{760 mmHgx15 L}{243 K} =58 mmHgx23 L$

$T2 =\frac{58 mmHgx23 L}{\frac{760 mmHgx15 L}{243 K}}$

T2= 28.44 K= -244.56 °C

Finally, the new temperature is -244.56 °C or 28.44 K.

Learn more about combined law equation:

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