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

`T_1=166.7K`

Step-by-step 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:

`(P_1)/(T_1) =(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=(T_2P_1)/(P_2)=(250K*1atm)/(1.5atm) \\\\T_1=166.7K`

Best regards.

Answer 2

Answer:

`T1=166.66K`

Step-by-step 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`

: