Need help with 1 A and 1 B

Choose the statement(s) that is/are true about the ratio \frac{C_p}{C_v} C p C v for a gas? (Ii) This ratio is the same for all

Blababa [14]

Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

Cp / Cv = 7 / 5 = 1.4

This statement is true.

(iv) It is false.

6 0

3 years ago

A. A light wave moves through glass (n=1.5) at an angle of 25°. What angle will it have when it moves from the glass into air (n

torisob [31]

PART A)

By Snell's law we know that

$n_1sin i = n_2 sin r$

here we know that

$n_1 = 1.5$

$i = 25 degree$

$n_2 = 1$

now from above equation we have

$1.5 sin25 = 1 sin r$

$r = 39.3 degree$

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}$

$d_i = 30 cm$

so image will form at 30 cm from mirror

it is virtual image and smaller in size

3 0

3 years ago

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. de

atroni [7]

Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c

5 0

2 years ago

An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This ca

dolphi86 [110]

Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

Let V be the potential difference of the battery.

The energy stored in the capacitor is given by

U = Q^2/2C ...(1)

Now the battery is disconnected, it means the charge is constant.

the separation between the plates is doubled.

The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.

C' = C/2

the new energy stored

U' = Q^2 / 2C'

U' = Q^2/C = 2 U

The energy stored in the capacitor is doubled the initial amount.

8 0

2 years ago

Someone help me in this question.

tekilochka [14]

Gradpoint ? And the answer is water.

4 0

2 years ago

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