Explanation:
Carbon dioxide is in a steel tank at 20°C, 10 liters and 1 atm.
Volume of the gas = 10 L
1) Initial Pressure of the gas [tex]P_1=1 atm[/tex]
Initial temperature of gas =[tex]T_1=20^oC=293 K[/tex]
Final pressure of the gas, when volume remains constant = [tex]P_2=?[/tex]
Final temperature of the gas [tex]T_2=100^oC=273 K[/tex]
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex] (Gay Luassac's law)
[tex]P_2=\frac{1 atm\times 373 k}{293 K}=1.27 atm[/tex]
1.27 atm is the pressure on the gas when the tank is heated to 100°C.
2) Initial Pressure of the gas [tex]P_1=1 atm[/tex]
Initial temperature of gas =[tex]T_1=20^oC=293 K[/tex]
Final pressure of the gas, when volume (215 mL) remains constant = [tex]P_2=1.5 atm[/tex]
Final temperature of the gas [tex]T_2=?[/tex]
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex] (Gay Luassac's law)
[tex]T_2=\frac{1.5 atm\times 293 K}{1 atm}=439.5 K=166.5 ^oC[/tex]
3)When the volume changes from 63.5 mL to 69.2 mL ,if the initial pressure was 735 mm Hg pressure.
Initial volume of the gas [tex]V_1=63.9 mL[/tex]
Initial pressure of gas =[tex]P_1=735 mmHg[/tex]
Final pressure of the gas, = [tex]P_2=?[/tex]
Final volume of the gas [tex]V_2=69.2 mL[/tex]
[tex]P_1\times V_1=P_2\times V_2[/tex] (Boyle's law)
[tex]P_2=\frac{63.5 mL\times 735 mmHg}{69.2 mL}=674.4 mmHg[/tex]
The 674.4 mmHg was the final pressure when gas occupies 69.2 mL of volume.
