Question

SO

25
7.) A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To
what temperature must the gas in the syringe be heated/cooled in order to have a volume of 435
mL at 2.50 atm?
A) 139 K
B) 572 K
C) 175 K
D) 466 K
E) 721 K
8.) What mass of NO2 is contained in a 13.0 L tank at 4.58 atm and 385 K?
A) 18.8 g
B) 86.7 g
C) 24.4 g
D) 53.1 g
E) 69.2 g

Answer 1

`\large \boxed{\text{E) 721 K; B) 86.7 g}}`

Step-by-step explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

`\begin{array}{rcl}(p_(1)V_(1))/(T_(1))& =&(p_(2)V_(2))/(T_(2))\\\\(1.88*285)/(355) &= &(2.50* 435)/(T_(2))\\\\1.509& = &(1088)/(T_(2))\\\\1.509T_(2) & = & 1088\\T_(2) & = & (1088)/(1.509)\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}`

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

`\begin{array}{rcl}pV & = & (mRT)/(M)\\\\4.58 * 13.0 & = & (m* 0.08206* 385)/(46.01)\\\\59.54 & = & 0.6867m\\m & = & (59.54)/(0.6867 )\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_(2)$ is $\large \boxed{\textbf{86.7 g}}$}`