a)21.4 L
b)19.6 L
c)32.7 L
2 Answers

If you’re told that the conditions are at STP, you automatically know that two of the quantities are T = 0ºC, or 273K and P = 101 kPa, or approximately 1 atm.
You’re given the value n in moles, so the only unknown in the equation VP = nRT is that for the volume V. Since you want the answer to result in L, it’s best to use the Gas Constant of R = 0.082057 L·atm·K^1·mol^1. Using the same units as those present in the Gas Constant ensure they properly cancel.
Thus,
P = 1 atm
T = 273 K
n = 0.875 mol
R = 0.082057 L·atm·K^1·mol^1
V = ?
Rearranging VP = nRT, we get:
V = nRT/P = (0.875)(0.082057)(273)/(1) = 19.6 L
Source(s): Ralph H. Petrucci et al. General Chemistry, Principles & Modern Applications, 9th Edition. Chapter 6, Gases; pages 178210. Person Prentice Hall: Upper Saddle RIver, New Jersey, 2007.ISBN 0121493302

at STP, the Vm=22.4
The Vm is found from equation PV=nRT
Vm=V/n = RT/P
where R is 0.082, T is 273K and P is 1atm
V=n x Vm
= 0.875 x 22.4
= 19.6 L (B)