Ask question

Ask question

All categories

2 years ago

14

(A) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pus

hed out through 10 cm against an external pressure of 1.0 atm. Calculate the work done by the system.
(B) A chemical reaction takes place in a container of cross-sectional area 50.0 cm2. As a result of the reaction, a piston is pushed out through 15 cm against an external pressure of 121 kPa. Calculate the work done by the system.

1 answer:

balandron [24]2 years ago

5 0

Answer:

(A) The work done by the system is -101.325J

(B) The workdone by the system is -90.75J

Explanation:

(A) Workdone = -PΔV

Given that A = 100cm2 = 0.01m2

distance d = 10cm = 0.1m

ΔV= Area × distance

ΔV= 0.01 ×0.1

ΔV = 0.001m3

P= external pressure = 1atm = 101325Pa

Workdone = -0.001 × 101325

W= - 101.325Pa m3

1Pam3 = 1J

Therefore W = - 101.325J

The work done on the system is -101.325J

(B) Workdone = -PΔV

Given that A = 50cm2 = 0.005m2

distance d = 15cm = 0.15m

ΔV= Area × distance

ΔV= 0.005×0.15

ΔV = 0.00075m3

P=121kPa = 121000Pa

W= - 121000 × 0.00075

W= -90.75Pa m3

1Pam3 = 1J

W = - 90.75J

The woekdone by the system is -90.75J

You might be interested in

What does the simplified model of the hall effect give for the density of free electrons in the unknown metal?

evablogger [386]

The simplified model of the hall effect proved that the current (electric) in metals are carried by electrons and not protons. The hall effect introduced the hall coefficient which is the ratio of the induced electric field to the current density x applied magnetic field. This coefficient is unique for each type of metal.

3 0

1 year ago

A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a SoaveRedlich-Kwong gas and with

Setler79 [48]

Explanation:

The given data is as follows.

Moles of propylene = 100 moles, $T_{i}$ = 300 K

$T_{f}$ = 800 K, $V_{i} = 2 m^{3}$

$V_{f} = 0.02 m^{3}$ , $C_{p}$ of propylene = 100 J/mol

Now, we assume the following assumptions:

Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.

W = $mC_{p} \Delta T$

$100 moles \times 100 J/mol K (800 - 300) K$

= $5 \times 10^{6} J$

= 5 MJ

Thus, we can conclude that a minimum of 5 MJ work is required without any friction.

3 0

1 year ago

Hector claims that a single sodium ion and a single oxygen ion do not bond together to form a stable binary ionic compound. Is h

levacccp [35]

Answer:

C

Explanation:

Looking at the periodic table, we can see that sodium is in group 1, so a sodium ion would be Na⁺, with a charge of +1. Oxygen is in group 16, so an oxygen ion would be O²⁻, with a charge of -2.

A compound formed only by a single sodium ion and a single oxygen ion would thus have a charge of -1, and in order to have a stable ionic compound its charge must be zero.

8 0

1 year ago

Read 2 more answers

A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago

NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including f

ELEN [110]

First, let's write down the balanced chemical reaction between the given reactants:

NO₂ + NO → N₂O + O₂

The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:

Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)

For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1

6 0

1 year ago