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 \(P_1=1 atm\)
Initial temperature of gas =\(T_1=20^oC=293 K\)
Final pressure of the gas, when volume remains constant = \(P_2=?\)
Final temperature of the gas \(T_2=100^oC=273 K\)
\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\) (Gay Luassac's law)
\(P_2=\frac{1 atm\times 373 k}{293 K}=1.27 atm\)
1.27 atm is the pressure on the gas when the tank is heated to 100°C.
2) Initial Pressure of the gas \(P_1=1 atm\)
Initial temperature of gas =\(T_1=20^oC=293 K\)
Final pressure of the gas, when volume (215 mL) remains constant = \(P_2=1.5 atm\)
Final temperature of the gas \(T_2=?\)
\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\) (Gay Luassac's law)
\(T_2=\frac{1.5 atm\times 293 K}{1 atm}=439.5 K=166.5 ^oC\)
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 \(V_1=63.9 mL\)
Initial pressure of gas =\(P_1=735 mmHg\)
Final pressure of the gas, = \(P_2=?\)
Final volume of the gas \(V_2=69.2 mL\)
\(P_1\times V_1=P_2\times V_2\) (Boyle's law)
\(P_2=\frac{63.5 mL\times 735 mmHg}{69.2 mL}=674.4 mmHg\)
The 674.4 mmHg was the final pressure when gas occupies 69.2 mL of volume.
