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The pressure in a natural gas tank is maintained at 2.20 atm. On a day when the temperature is-15 °C, the volume of gas in the tank is 3.25 x 10^3 m^3. What is the volume of the same quantity of gas on a day when the temperature is 31 °C?

User Kakamotobi
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2 Answers

4 votes

Final answer:

To determine the volume of gas at a different temperature while keeping pressure constant, Charles's Law is used. The volume of the gas on a day when the temperature is 31 °C is approximately 3.83 x 10^3 m^3, given an initial volume of 3.25 x 10^3 m^3 and temperature change from -15 °C to 31 °C.

Step-by-step explanation:

The question pertains to the behavior of gases under different temperatures while keeping pressure in a natural gas tank constant, which is a typical Chemistry problem involving the Ideal Gas Law. To find the volume of the same quantity of gas under different temperature conditions, we use Charles's Law, which states that if the pressure of a gas is held constant, its volume is directly proportional to the absolute temperature. We convert temperatures from Celsius to Kelvin by adding 273.15 and then apply the formula:

V1/T1 = V2/T2

Where:

V1 = initial volume = 3.25 x 10^3 m^3

T1 = initial temperature in Kelvin = -15 °C + 273.15 = 258.15 K

T2 = final temperature in Kelvin = 31 °C + 21.15 = 304.15 K

V2 = final volume (the value we want to find)

We then rearrange terms to solve for V2:

V2 = V1 * (T2/T1)

V2 = 3.25 x 10^3 m^3 * (304.15 K / 258.15 K)

V2 = 3.25 x 10^3 m^3 * 1.178 ≈ 3.83 x 10^3 m^3

Thus, the volume of the gas on a day when the temperature is 31 °C is approximately 3.83 x 10^3 m^3.

User Mkarasek
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5 votes

Final answer:

To find the volume of the gas on a day when the temperature is 31 °C, we can use the combined gas law equation: P1V1/T1 = P2V2/T2. Substituting the given values and solving for P2, we find that the volume of the gas is 8.33 atm.

Step-by-step explanation:

To find the volume of the gas on a day when the temperature is 31 °C, we can use the combined gas law equation:

P1V1/T1 = P2V2/T2

Given:

  • P1 = 2.20 atm
  • V1 = 3.25 x 10^3 m^3
  • T1 = -15 °C + 273.15 = 258.15 K
  • T2 = 31 °C + 273.15 = 304.15 K

Substituting these values into the equation:

(2.20 atm)(3.25 x 10^3 m^3)/(258.15 K) = P2(unknown volume)/(304.15 K)

Solving for P2:

P2 = (2.20 atm)(3.25 x 10^3 m^3)(304.15 K)/(258.15 K) = 8.33 atm

Therefore, the volume of the gas on a day when the temperature is 31 °C is 8.33 atm.

User Taketwo
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