Question

A gas sample occupies 15.44L at 42.0°C. What is the pressure in kilopascals, given that there are 6.73mol of a gas in the sample?

Answer 1

Use Ideal Gas Law

$PV=nRT$

$V= 15.44 dm^3(\frac{1 m^3}{1000 dm3})= 15.44\times 10^{-3}m^3$

$T(K) = 273+42=315K$

$P=\frac{nRT}{V}=\frac{6.73\times 8.314\times 315 }{15.44\times 10^{-3}}=1141533 Pa = 1141kPa = 1140kPa$ (3 s.f.)