Question

A sample of oxygen gas in one container has a volume of 20.0mililiter at 297 K and 101.3 kPa. The entire sample is transferred to another container where the temperature is 283 K and the pressure is 94.6 kPa. Which is correct numerical setup for calculating the new volume of this sample of oxygen gas
Show work please

Answer 1

Answer: $V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}$

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,:

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

$V_2=\frac{P_1V_1T_2}{T_1P_2}$

where,

$P_1$ = initial pressure of gas = 101.3 kPa

$P_2$ = final pressure of gas = 94.6 kPa

$V_1$ = initial volume of gas = 20.0 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $297K$

$T_2$ = final temperature of gas = $283K$

Now put all the given values in the above equation, we get the final volume of gas.

$V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}$

$V_2=20.4ml$

Thus the correct numerical setup for calculating the new volume is $\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}$