Question

A certain amount of gas at 25 C and at a pressure of 0.800 atm is contained in a glass vessel. Suppose that the vessel can withstand a pressure of 2.00 atm. How high can you raise the temperature of the gas without bursting the vessel?

a. P1 / T1 = P2 / T2

b. so .8/298=2/T2

c. which equals 745 K

Answer 1

Answer : 745 K high you can raise the temperature of the gas without bursting the vessel.

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

`P\propto T`

or,

`(P_1)/(T_1)=(P_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 0.800 atm

`P_2` = final pressure of gas = 2.00 atm

`T_1` = initial temperature of gas =
`25^oC=273+25=298K`

`T_2` = final temperature of gas = ?

Now put all the given values in the above equation, we get:

`(0.800atm)/(298K)=(2.00atm)/(T_2)`

`T_2=745K`

Therefore, the final temperature of the gas is 745 K

Answer 2

=745 Kelvin

Step-by-step explanation:

• According to the pressure law the relationship between the pressure and absolute temperature of a gas at a constant volume is directly proportional.
• An increase in temperature causes a corresponding increase in pressure at a constant volume.

That is;

`P\alpha T`

Mathematically;

`P = kT\\(Constant)k =(P)/(T) \\`

When there are two varying conditions;

`(P1)/(T1) =(P2)/(T2)`

In this case;

P1 = 0.8 atm

T1 = 25+273 = 298 K

P2 =2.00 atm

T2= ?

Therefore;

`T2= (P2T1)/(P1)`

`T2= ((2)(298))/(0.8)`

`= 745 K`

Therefore, the temperature can be raised up to 745 kelvin without bursting vessel.