Which one of the following gases occupies the smallest volume at STP?

A.4g helium
B.14g nitrogen
C.34g ammonia
D.32g oxygen

4g helium,14g nitrogen, 34g ammonia,32g oxygen

To determine which gas occupies the smallest volume at STP (Standard Temperature and Pressure), we need to understand the relationship between gas volume and molar mass.

The ideal gas law equation is represented as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

First, we need to convert the given masses of each gas into moles. We can do this by dividing the mass of each gas by its molar mass. The molar mass of helium (He) is approximately 4 g/mol, of nitrogen (N2) is approximately 28 g/mol, of ammonia (NH3) is approximately 17 g/mol, and of oxygen (O2) is approximately 32 g/mol.

Calculating the number of moles for each gas:
A. 4g He / 4 g/mol = 1 mol He
B. 14g N2 / 28 g/mol = 0.5 mol N2
C. 34g NH3 / 17 g/mol = 2 mol NH3
D. 32g O2 / 32 g/mol = 1 mol O2

Now, let's consider the ideal gas law equation, assuming the pressure and temperature are constant.

At STP, we have a fixed pressure of 1 atmosphere (atm) and a temperature of 273 Kelvin (K). Since the pressure and temperature are constant, we can simplify the equation to V = nR, where V is the volume occupied by each gas, and R is the gas constant.

The gas constant, R, is the same for all gases under the same conditions and has a value of approximately 0.0821 L·atm/(mol·K).

Calculating the volume for each gas:
A. V = 1 mol He * 0.0821 L·atm/(mol·K) = 0.0821 L
B. V = 0.5 mol N2 * 0.0821 L·atm/(mol·K) = 0.04105 L
C. V = 2 mol NH3 * 0.0821 L·atm/(mol·K) = 0.1642 L
D. V = 1 mol O2 * 0.0821 L·atm/(mol·K) = 0.0821 L

Comparing the volumes obtained for each gas, we can see that the gas with the smallest volume at STP is option B: 14g nitrogen. It occupies a volume of approximately 0.04105 liters.