Question

One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is heated at constant volume until the pressure triples, what is the fi nal temperature? (b) If the gas is heated so that both the pressure and volume are doubled, what is the fi nal temperature?

Answer 1

Answer: a) 900 K

b) 1200 K

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 6.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K    $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas  = 1

$V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L$

a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

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

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?$

Putting values in above equation, we get:

$\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K$

The final temperature is 900 K

b) The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 6.00 atm

$P_2$ = final pressure of gas = $2\times 6.00atm=12.0atm$

$V_1$ = initial volume of gas = 4.10 L

$V_2$ = final volume of gas =  $2\times 4.10 L=8.20L$

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas =?

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

$\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}$

$T_2=1200K$

The final temperature is 1200 K