Question

What is the pressure in kPa of 2 moles of chlorine gas at a temperature of 5 degrees C
and a volume of 8 dm3?

Answer 1

Answer:

$P=5.8x10^2kPa$

Explanation:

Hello.

In this case, since chlorine gas is assumed to behave ideally, we can use the ideal gas equation in order to compute the required pressure:

$PV=nRT$

In such a way, since the temperature must be in kelvins and the volume in liters for the universal gas constant:

$T=5+273.15=278.15K\\\\V=8dm^3*\frac{1L}{1dm^3}=8L$

Thus, we first compute the pressure in atm:

$P=\frac{nRT}{V}= \frac{2mol*0.082\frac{atm*L}{mol*K}*278.15K}{8L}\\\\P=5.7atm$

Thus, the pressure in kPa turns out:

$P=5.7atm*\frac{101.325kPa}{1atm}\\\\P=5.8x10^2kPa$