Question

A 600.0 mL sample of krypton is held at standard conditions. What volume does the gas occupy when the pressure is tripled and the temperature doubles

Answer 1

Answer: The volume when the pressure and temperature has changed is 400 mL

Explanation:

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

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

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=P_1\\V_1=600.00mL\\T_1=T_1\\P_2=3P_1\\V_2=?\\T_2=2T_1$

Putting values in above equation, we get:

$\frac{P_1\times 600mL}{T_1}=\frac{3P_1\times V_2}{2T_1}\\\\V_2=\frac{600\times P_1\times 2T_1}{T_1\times 3P_1}=400mL$

Hence, the volume when the pressure and temperature has changed is 400 mL