A gas is confined to a steel tank with a fixed volume. At 293 Kelvin (K), the gas exerts a pressure of 5.0 atmosphere (atm). After heating the tank, the pressure of the gas increases to 10.0 atm. What is the temperature of the heated gas?*2 puntos586 K326 K926 K357 K
Question
A gas is confined to a steel tank with a fixed volume. At 293 Kelvin (K), the gas exerts a pressure of 5.0 atmosphere (atm). After heating the tank, the pressure of the gas increases to 10.0 atm. What is the temperature of the heated gas?*2 puntos586 K326 K926 K357 K
Solution
To solve this problem, we can use the Gay-Lussac's Law which states that the pressure of a gas is directly proportional to its temperature if the volume and the amount of gas are kept constant. The formula for Gay-Lussac's Law is P1/T1 = P2/T2.
Here, P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature.
Given in the problem: P1 = 5.0 atm T1 = 293 K P2 = 10.0 atm T2 = ?
We can substitute these values into the formula and solve for T2:
(5.0 atm / 293 K) = (10.0 atm / T2)
Cross-multiplying gives us:
5.0 atm * T2 = 10.0 atm * 293 K
Solving for T2 gives us:
T2 = (10.0 atm * 293 K) / 5.0 atm
T2 = 586 K
So, the temperature of the heated gas is 586 Kelvin.
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