Question

An ideal gas originally at 0.85 atm and 66°C was allowed to expand until its final volume, pressure, and temperature were 94 mL, 0.60 atm, and 43°C, respectively. What was its initial volume?

Answer 1

Answer : The initial volume was, 71.2 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:

$P_1=0.85atm\\V_1=?\\T_1=66^oC=[66+273]K=339K\\P_2=0.60atm\\V_2=94mL\\T_2=43^oC=[43+273]K=316K$

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

$\frac{0.85atm\times V_1}{339K}=\frac{0.60atm\times 94mL}{316K}$

$V_1=71.2mL$

Therefore, the initial volume was, 71.2 mL