# Question 942bc

Nov 26, 2014

1) Convert 10.3 °C to Kelvin and you get 273 + 10.3= 283.3 K . At this temperature the volume of the gas is 127 L.

2) At Temperature 103.0 °C or 273 + 103 = 376 K this temperature the volume of the gas is ${V}_{2}$ mL.

Remember that you have to plug into the equation in a very specific way. The temperatures and volumes come in connected pairs and you must put them in the proper place.

Using Charles law equation;

${V}_{1}$ / ${T}_{1}$ = ${V}_{2}$ / ${T}_{2}$

${V}_{1}$ = 127 L , ${T}_{1}$ = 283.3 K

${V}_{2}$ = ? , ${T}_{2}$ = 376 K

plug in the values;

127 L / 283.3 K = ${V}_{2}$ / 376K

Cross-multiply and divide:

127 L x 376 K = ${V}_{2}$ x 283.3 K

47752 L K = ${V}_{2}$ x 283.3 K

${V}_{2}$ = 47752 L K / 283.3 K

${V}_{2}$ = 168.5 L = 169 L due to significant figures.

Nov 27, 2014

Charles' law states that when the pressure of a gas is kept constant, the temperature and volume have a direct relationship. This means that if the temperature is increased, the volume will also increase.

The equation for Charles' law is:

${\text{V"_1/"T}}_{1}$ = ${\text{V"_2/"T}}_{2}$

For the gas laws, temperature must be in Kelvins. $\text{K}$ = $\text{C"^"o} + 273$.

To solve problems involving Charles' law, we cross multiply and isolate the missing variable. Then plug in the known values and solve.

Since another contributor already showed how to solve the first problem, this answer will focus on the other three.

Problem 2. Solve for ${\text{V}}_{1}$.

Cross multiply and solve for "V_1#.

$\text{V"_1}$ = $\text{V"_2"T"_1"/T"_2}$ = $\text{(185L)(129K)"/"325K}$ =$\text{73.4L}$

Problem 3. Solve for $\text{T"_2}$

Convert ${\text{T}}_{1}$ to Kelvins.

${\text{T}}_{1}$ = $\text{15.5"^"o""C}$ + $\text{273K}$ = $\text{288.5K}$

Cross multiply and solve for ${\text{T}}_{2}$.

${\text{T}}_{2}$ = $\text{V"_2"T"_1"/V"_1}$ = $\text{(2.43L)(288.5K)"/"2.04L}$ = $\text{344K}$

$\text{T"_2" in Celsius}$ = $\text{344K" - 273 = "71.0"^"o""C}$.

Problem 4. Solve for ${\text{T}}_{1}$.

Convert ${\text{T}}_{1}$ to Kelvins.

${\text{T}}_{2}$ = $\text{11.1"^"o""C}$ + $\text{273}$ = $\text{284.1K}$

Cross multiply and solve for ${\text{T}}_{1}$.

${\text{T}}_{1}$ = ${\text{V"_1"T"_2"/V}}_{2}$= $\text{(14.8cm"^3")(284.1K)/""18.9cm"^3}$ = $\text{222K}$