Ask question

Ask question

All categories

3 years ago

12

Two pounds of water vapor at 30 psia fill the 4-ft3 left chamber of a partitioned system. The right chamber has twice the volume

of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 40oF.

1 answer:

fomenos3 years ago

6 0

Answer:

3.38atm

Explanation:

Using data from the steam table we have that

Moles of water vapour = 907.19 / 18

= 50.4 moles

So

p1 = 30 psi = 30 x 0.68 = 2.04 atm

v1 = 4ft³= 113.2 L

Then from

PV= nRT

Then to find T we use

T1 = p1 V1 / n R

= 2.04 x 113.2 / 50.4 x 0.0821

= 55.8 K

Then to find volume two

v2 = 2v1 + v1

So

3 v1 = 339.6 K

The pressure two we use

P2 = n R T2 / V2

= 50.4 x 0.0821 x 277.6 / 339.6

So we have

= 3.38 atm =

You might be interested in

What is the kinetic energy of a 50-kg child running to catch the school bus at

Vinvika [58]

Answer:

Option C

100 J

Explanation:

Kinetic energy, KE is given by

$KE=0.5mv^{2}$ where m is the mass and v is the velocity

Substituting 50 Kg for mass, m and 2 m/s for velocity v then we obtain

$KE=0.5*50*2^{2}=100 J$

Therefore, the child's kinetic energy is equivalent to 100 J

6 0

3 years ago

Read 2 more answers

What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?

statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000 x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J

Kinetic energy is given as;

K.E = ¹/₂mv²

where;

v is speed of the electron

$K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} = (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} = (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s$

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

6 0

3 years ago

Compare and contrast repetition and replication in a science investigation

Bas_tet [7]

Repetition occurs when multiple sets of measurements are made during one scientific investigation. Replication occurs when a scientific investigation is <span>reproduced by another person.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>

3 0

3 years ago

Which of the following statements describes a compound machine?

ASHA 777 [7]

From the word compound, the compound machine is already a combination of two or more types of simple machine. Thus, the answer is letter C. Because of its complexity, it is able to perform several other functions than a simple one.

6 0

2 years ago

Read 2 more answers

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago