Question

The pressure of a gas in a cylinder when it is heated to a temperature of 250k is 1.5 atm. What was the initial temperature of the gas if it’s initial pressure was 1 atm?

Answer 1

Answer:

$T1=166.66K$

Explanation:

According to Gay-Lussac’s law simply states that the ratio of the initial pressure and temperature is equal to the ratio of the final pressure and temperature for a gas of a fixed provided that the mass is kept at a constant volume.

Given:

Initial pressure, P1 = 1 atm

Final pressure, P2 = 1.5 atm

Final temperature, T2 = 250 K

The law can be applied using the below formula

P1T2 = P2T1

Then,

T1 = (P1T2)/P1 = (1*250)/(1.5) = 166.66 Kelvin.

$T1=166.66K$

:

Answer 2

Answer:

$T_1=166.7K$

Explanation:

Hello,

In this case, by applying the Gay-Lussac's law which help us to understand the pressure-temperature gas behavior via a directly proportional relationship:

$\frac{P_1}{T_1} =\frac{P_2}{T_2}$

In such a way, as we are asked to compute the initial temperature knowing the initial pressure and final both temperature and pressure, so we solve for it:

$T_1=\frac{T_2P_1}{P_2}=\frac{250K*1atm}{1.5atm} \\\\T_1=166.7K$

Best regards.