Chemistry Homework: Solving Gas Laws Problems - Chapter 5, Semester 1

This document provides detailed solutions to a chemistry homework assignment focused on gas laws. The assignment covers various concepts including pressure conversions (torr to atm), calculations involving Boyle's Law, and the application of the ideal gas law. It also includes problems related to partial pressures, volume calculations, and effusion rates of different gases. Furthermore, the solutions incorporate the use of manometers and the interpretation of their readings. The document also addresses calculations involving the mass and volume of gases produced in chemical reactions, and the impact of pressure changes on gas behavior, including an extra credit question on piston movement.

Q 1
1 torr = 0.0013157 atm
27.3 torr= 27.3 * 0.0013157
= 0.03592 atm
The pressure of the sample = 1.07 – 0.03592
= 1.034 atm
Q2
P1 v1= p2 v2
6*1 = v2 * 12.5
V2=0.48 L
Q3
P
t = p 1
t 1
7.1
273+27 = p 1
273+5
P1= 6.579 atm
Q4
Added volume = 2.5∗50
134.2

= 0.9314
New vol= 0.9314 + 2.5
= 3.431 L
Q5
Assuming the density of oxygen = 1.23 g/l
125L ≈ 125 x 1.23 g/l
= 153.75 g of O2
Since the ration of reactions anre products are 1:1:1:1:1
Then 153.75 g of NClO3 are needed
Q6
P1 v1
T 1 = p2
v2
T 2
755∗100
293 = 195∗V
221
V= 292.03 L
Q7
From the ratio of the Zn and He (1:1), the mass of He gas generated can be calculated;
Mass= 10g *1

= 10g
From density of gas = 1.23g/l the volume can be determine
Vol= 10 g
1.23 g l−1
= 8.13 L
Q8
a. ,
757mm Hg = 0.9961 atm
Using
PV= nTR
0.9961 * 1L = n * 298 K * 0.08206
n= 0.0407
molar mass = 0.65g/ 0.0407 moles
= 15g/mol
b. HCl was the gas
Q9

(P1 v1)+( p2 v2)=p3∗v3
(2*3.25)+(3*1.6) = p3 * 5
P3= 1.932 atm
Q10
R 1
R 2 = √ ( m2
m1 )
Mass of C2H2 = 26g
a. For N2 mass = 28
Therefore ;
Rc 2 H 2
RN2
= √ ( m2
m1 )
= 1.03 higher diffusion rate
b. O2 ( mass = 16 g)
Rc 2 H 2
RO2
= √ ( 32
26 )
= 1.11 hence higher effusion
c. Cl2 (mass = 34g)

Rc 2 H 2
RCl2
= √ ( 70
26 )
= 1.643 and therefore higher effusion rate
d. CH4 (mass= 16g )
Rc 2 H 2
RCH4
= √ (16
26 )
= 0.78 hence lower effusion
e. CO2 (mass = 44g )
Rc 2 H 2
RCO2
= √ ( 44
26 )
= 1.3 therefore faster
Extra credit
Decreasing the pressure outside the container will make the piston move from the
base.