Question

activity c (continued from previous page)

1. compare: the accepted value for r is 0.08206 l·atm/k·mol or 8.314 l·kpa/k·mol, depending on the unit of pressure used. (your answer may differ slightly due to rounding.)

how close was your calculation?

1. synthesize: the ideal gas law is an equation relating p, v, r, n, and t. rewrite the formula you found in question 3a so that p and v are on one side and r, n, and t are on the other. show your work.

1. discover: it is important to have a baseline set of conditions to serve as a reference point. standard temperature and pressure (stp) is defined as 1 atmosphere (atm) or 101.325 kilopascals (kpa) of pressure at 273 k (0 °c). stp reflects normal atmospheric conditions at sea level.

a. use the gizmo to find the volume of 1 mole of gas at stp. (you will need to manually enter the temperature.) what value did you find?

b. choose a different gas. does the volume change?

1. calculate: use the ideal gas law (pv = nrt) to solve the following. show work for each problem. then use the gizmo to check your answer.

a. what is the volume of 0.5 moles of gas at stp?

v =

b. how much pressure would 0.8 moles of a gas at 370 k exert if it occupied 17.3 l of space?

p =

c. how much h₂ gas is necessary to exert a pressure of 1.4 atm at 430 k if occupying a volume of 15.1 l?

Explanation:

5. Synthesize (Ideal Gas Law Manipulation)

Step1: Recall the Ideal Gas Law

The ideal gas law is \( PV = nRT \) (assuming the formula from 3A is equivalent to this, as it relates \( P, V, n, R, T \)).

Step2: Ensure \( P \) and \( V \) on One Side

The equation \( PV = nRT \) already has \( P \) and \( V \) multiplied on the left - hand side, and \( n, R, T \) on the right - hand side (since \( nRT=n\times R\times T \)). So no further algebraic manipulation is needed beyond recognizing the standard ideal gas law form. If we assume the formula from 3A was derived to be equivalent to the ideal gas law, then the rearrangement is straightforward.

Step1: Identify STP Conditions and Formula

At STP, \( P = 1\space atm \), \( T=273\space K \), \( n = 0.5\space mol \), and \( R = 0.08206\space L\cdot atm/K\cdot mol \). We use the ideal gas law \( PV=nRT \) and solve for \( V \): \( V=\frac{nRT}{P} \)

Step2: Substitute Values

Substitute \( n = 0.5\space mol \), \( R = 0.08206\space L\cdot atm/K\cdot mol \), \( T = 273\space K \), and \( P = 1\space atm \) into the formula:
\( V=\frac{0.5\space mol\times0.08206\space L\cdot atm/K\cdot mol\times273\space K}{1\space atm} \)
First, calculate the numerator: \( 0.5\times0.08206\times273=0.5\times22.40238 = 11.20119\space L \)

Step1: Rearrange Ideal Gas Law for \( P \)

From \( PV=nRT \), solve for \( P \): \( P=\frac{nRT}{V} \)

Step2: Substitute Values

We have \( n = 0.8\space mol \), \( R = 0.08206\space L\cdot atm/K\cdot mol \), \( T = 370\space K \), and \( V = 17.3\space L \)
Substitute into the formula: \( P=\frac{0.8\space mol\times0.08206\space L\cdot atm/K\cdot mol\times370\space K}{17.3\space L} \)
First, calculate the numerator: \( 0.8\times0.08206\times370=0.8\times30.3622 = 24.28976 \)
Then divide by \( V \): \( P=\frac{24.28976}{17.3}\approx1.404\space atm \) (or using \( R = 8.314\space L\cdot kPa/K\cdot mol \), we would convert units appropriately, but using \( R = 0.08206\space L\cdot atm/K\cdot mol \) is consistent with STP pressure in atm)

Answer:

The formula with \( P \) and \( V \) on one side and \( R \), \( n \), and \( T \) on the other is \( PV = nRT \)

7A. Calculate Volume at STP