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The temperature of a sample of gas in a steel container at 25.0 kPa starts at -50 C and decreases by a factor of three. What is the final pressure inside the tank?

2 Answers

7 votes

Answer:

So if pressure of a gas is increased by 25%, the volume of a gas is decreased by 25%.

Step-by-step explanation:

User Avaq
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Answer: The final pressure inside the tank is 8.41 kPa.

Explanation: We can use the combined gas law to solve this problem, which relates the pressure, volume, and temperature of a gas:

(P1V1)/T1 = (P2V2)/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

We are given P1 = 25.0 kPa, T1 = -50 C = 223 K, and V1 is unknown. We also know that the temperature decreases by a factor of three, so T2 = T1/3 = 223/3 K.

To find V2, we need to assume that the steel container is rigid and its volume remains constant. Therefore, V1 = V2, and we can cancel out the volume from the equation:

P1/T1 = P2/T2

Substituting the values, we get:

P2 = P1 * T2 / T1 = 25.0 * (223/3) / 223 = 8.41 kPa

Therefore, the final pressure inside the tank is 8.41 kPa.

User FredyC
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