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 Kelvin

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;

$P = kT\\(Constant)k =\frac{P}{T}$

When there are two varying conditions;

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case;

P1 = 0.8 atm

T1 = 25+273 = 298 K

P2 =2.00 atm

T2= ?

Therefore;

$T2= \frac{P2T1}{P1}$

$T2= \frac{(2)(298)}{0.8}$

$= 745 K$

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

Answer 2

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,

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

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:

$\frac{0.800atm}{298K}=\frac{2.00atm}{T_2}$

$T_2=745K$

Therefore, the final temperature of the gas is 745 K