For many purposes we can treat dimethyl ether C2H6O as an ideal gas at temperatures above its boiling point of −24.°C.Suppose the pressure on a 9.0m3 sample of dimethyl ether gas at 19.0°C is tripled.Is it possible to change the temperature of the dimethyl ether at the same time such that the volume of the gas doesn't change? yesnoIf you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C. °C
Question
For many purposes we can treat dimethyl ether C2H6O as an ideal gas at temperatures above its boiling point of −24.°C.Suppose the pressure on a 9.0m3 sample of dimethyl ether gas at 19.0°C is tripled.Is it possible to change the temperature of the dimethyl ether at the same time such that the volume of the gas doesn't change? yesnoIf you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C. °C
Solution
Yes, it is possible to change the temperature of the dimethyl ether at the same time such that the volume of the gas doesn't change. This is because of the ideal gas 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.
To calculate the new temperature, we can use the formula for the ideal gas law, which is P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
First, we need to convert the temperatures from Celsius to Kelvin, because the ideal gas law requires temperatures to be in Kelvin. The conversion is T(K) = T(°C) + 273.15.
So, the initial temperature T1 is 19.0°C + 273.15 = 292.15 K.
The final pressure P2 is three times the initial pressure P1, so P2 = 3P1.
Substituting these values into the ideal gas law gives us:
P1/292.15 K = 3P1/T2
Solving for T2 gives us:
T2 = 3 * 292.15 K = 876.45 K
Finally, we convert this back to Celsius:
T2 = 876.45 K - 273.15 = 603.3°C
Rounding to the nearest degree gives us a final temperature of 603°C.
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