If the moles and volume of a gas are held constant, dropping the temperature from 40 degrees Celsius to 20 degrees Celsius causes would cause an initial pressure of 3 atmospheres to drop to 1.5 atmospheres.

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In this case, as the moles are held constant, one uses the Gay-Lussac's law in order to relate both the temperature and the pressure at the beginning and at the end of the experiment via the following relationship:

\(P_1T_2=P_2T_1\)

By solving for the pressure at the second (final) state we can substantiate whether the 1.5 atm are correct as shown below:

\(P_2=\frac{P_1T_2}{T_1}=\frac{3atm*293.15K}{313.15K}= 2.8 atm\)

Therefore, even when the pressure is correctly decreased, the given value is wrong as 2.8 atm is the correct final pressure, in this manner, the statement is false.

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If the moles and volume of a gas are held constant, dropping the temperature from 40 degrees Celsius to 20 degrees Celsius causes would cause an initial pressure of 3 atmospheres to drop to 1.5 atmospheres.T/F

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If the moles and volume of a gas are held constant, dropping the temperature from 40 degrees Celsius to 20 degrees Celsius causes would cause an initial pressure of 3 atmospheres to drop to 1.5 atmospheres.

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