A certain volume of a gas at 300K is heated such that its volume is 3X the original amount and its pressure is 4.25X the original amount. Find the new temp
Question
A certain volume of a gas at 300K is heated such that its volume is 3X the original amount and its pressure is 4.25X the original amount. Find the new temp.
Solution
To solve this problem, we can use the ideal gas law, which states that the product of the pressure and volume of a gas is directly proportional to the temperature of the gas. Mathematically, this is represented as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
In this case, we are not given the number of moles or the gas constant, but we can assume that they remain constant throughout the problem. This allows us to simplify the ideal gas law to the form P1V1/T1 = P2V2/T2, where the subscripts 1 and 2 refer to the initial and final states of the gas, respectively.
Given that the final volume is 3 times the initial volume (V2 = 3V1) and the final pressure is 4.25 times the initial pressure (P2 = 4.25P1), we can substitute these values into the simplified ideal gas law to find the final temperature:
P1V1/T1 = P2V2/T2 P1V1/T1 = (4.25P1)(3V1)/T2 1/T1 = 12.75/T2
Solving for T2 gives us:
T2 = T1 * 12.75
Given that the initial temperature T1 is 300K, the final temperature T2 is:
T2 = 300K * 12.75 = 3825K
So, the new temperature of the gas is 3825K.
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