Question

The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens to the pressure of the gas?

Answer 1

Answer:

• The pressure of a gas increases four times, when the absolute temperature is increased four times, while maintaining the volume constant.

Explanation:

The expression that rules the change of temperature of ideal gases at constant volume is the Law of Gay-Lussac: pressure and temperature of gases are directly related. In the form of equations that is:

• P / T = constant

• P₁ / T₁ = P₂ / T₂ .......... [equation 1]

The question states that the absolute temperature is increased four times, the you can write that as T₂ = 4 × T₁, and substitute in the equation 1 to obtain:

• P₁ / T₁ = P₂ / (4 × T₁)

Simplify:

• P₂ = P₁ × 4 × T₁ / T₁ = P₁ × 4

That proves that the pressure also increases four times, when the absolute temperature is increased four times, while maintaining the volume constant.

Answer 2

Answer: The pressure will also increase 4 times.

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=p\\T_1=t\\P_2=?=\\T_2=4t$

Putting values in above equation, we get:

$\frac{p}{t}=\frac{P_2}{4t}\\\\P_2=4p$

Hence, the pressure will also increase 4 times.