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

User Zouying
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2 Answers

3 votes

Answer: The pressure will also increase 4 times.

Step-by-step 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,


(P_1)/(T_1)=(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:


(p)/(t)=(P_2)/(4t)\\\\P_2=4p

Hence, the pressure will also increase 4 times.

User Xavierzhao
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6 votes

Answer:

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

Step-by-step 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.

User Arun Rana
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