Question

name: james wheat
hour: 1 2 3 4 5
the ideal gas law
use the ideal gas law, $pv=nrt$, to solve the following problems:

1. if i have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature?

given
$p=5.6 atm$
$v=12 l$
$r=r$
$n=4$
$t=t$
work
$(5.6)$
answer:
(k)
(204.83

1. if i have an unknown quantity of gas at a pressure of 121.6 kpa, a volume of 31 liters, and a temperature of $87 ^\circ c$, how many moles of gas do i have?

given
$p=p$
$v=31 l$
$r=8.314$
$n=n$
$t=87$
work

answer:
(mol)
(1.26

1. if i contain 3 moles of gas in a container with a volume of 60 liters and at a temperature of 400 k, what is the pressure (in atm) inside the container?

given
$p=p$
$v=60 l$
work

answer:
(atm)
(1.64

1. if i have 7.7 moles of gas at a pressure of 68.4 mm hg and at a temperature of $56 ^\circ c$, what is the volume of the container that the gas is in?

given
$p=68.4$
$t=56 ^\circ c$ $329 k$
$v=v$
$n=7.7 mol$
$r=62.4$
work
$(68.4)v=(7.7)(62.4)(329)$
$\frac{}{68.4}$
answer:
(2311.08 l)
$2311.08 l$

Explanation:

Problem 1:

Step1: Recall Ideal Gas Law

The ideal gas law is \( PV = nRT \). We need to solve for \( T \), so rearrange the formula to \( T=\frac{PV}{nR} \).
Given: \( P = 5.6\ \text{atm} \), \( V = 12\ \text{L} \), \( n = 4\ \text{mol} \), \( R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}} \) (standard gas constant for these units).

Step2: Substitute Values

Substitute the values into the formula: \( T=\frac{(5.6\ \text{atm})(12\ \text{L})}{(4\ \text{mol})(0.0821\ \frac{\text{L·atm}}{\text{mol·K}})} \)

Step3: Calculate Numerator and Denominator

Numerator: \( (5.6)(12)=67.2 \)
Denominator: \( (4)(0.0821)=0.3284 \)

Step4: Divide to Find \( T \)

\( T=\frac{67.2}{0.3284}\approx204.6\ \text{K} \) (close to the given 204.83, slight difference due to R precision)

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

From \( PV = nRT \), solve for \( n \): \( n=\frac{PV}{RT} \). First, convert temperature to Kelvin: \( T = 87 + 273 = 360\ \text{K} \), \( P = 121.6\ \text{kPa} \), \( V = 31\ \text{L} \), \( R = 8.314\ \frac{\text{L·kPa}}{\text{mol·K}} \) (for kPa units).

Step2: Substitute Values

\( n=\frac{(121.6\ \text{kPa})(31\ \text{L})}{(8.314\ \frac{\text{L·kPa}}{\text{mol·K}})(360\ \text{K})} \)

Step3: Calculate Numerator and Denominator

Numerator: \( (121.6)(31)=3769.6 \)
Denominator: \( (8.314)(360)=2993.04 \)

Step4: Divide to Find \( n \)

\( n=\frac{3769.6}{2993.04}\approx1.26\ \text{mol} \) (matches the given 1.26)

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

From \( PV = nRT \), \( P=\frac{nRT}{V} \). Given: \( n = 3\ \text{mol} \), \( V = 60\ \text{L} \), \( T = 400\ \text{K} \), \( R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}} \)

Step2: Substitute Values

\( P=\frac{(3\ \text{mol})(0.0821\ \frac{\text{L·atm}}{\text{mol·K}})(400\ \text{K})}{60\ \text{L}} \)

Step3: Calculate Numerator

\( (3)(0.0821)(400)=98.52 \)

Step4: Divide by Volume

\( P=\frac{98.52}{60}=1.642\ \text{atm} \approx1.64\ \text{atm} \) (matches the given 1.64)

Answer:

\( \approx204.6\ \text{K} \)

Problem 2: