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2 years ago

What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?

1 answer:

4 0

As you mentioned, we will use Equipartition Theorem.
H2 has 5 degrees of freedom; 3 translations and 2 rotation
Therefore:
Internal energy = (5/2) nRT
You just substitute in the equation with the values of R and T and calculate the internal energy as follows:
Internal energy = (5/2) x 2 x 8.314 x 308 = 32.0089 x 10^3 J

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Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi

Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

$\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}$

here we know that

$\mu_1 = 1.473$

$h_o = 5 cm$

$\mu_2 = 1$

now by above formula

$\frac{1.473}{5} = \frac{1}{h_i}$

$h_i = 3.39 cm$

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0

2 years ago

Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.

dimulka [17.4K]

Answer:

The distance of separation is $d = 0.092 \ m$

Explanation:

The mass of the each ball is $m= 10 g = 0.01 \ kg$

The negative charge on each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

$F = \frac{kq_1 * q_2}{d}$

=> $m* g = \frac{kq_1 * q_2}{d}$

here k the the coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

$0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

$d = 0.092 \ m$

5 0

2 years ago

If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?

Gnom [1K]

The average current density in the wire is given by:

$J=\frac{I}{A}$

where I is the current intensity and A is the cross-sectional area of the wire.

The cross-sectional area of the wire is given by:

$A=\pi r^2$

where r is the radius of the wire. In this problem, $r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m$ , so the cross-sectional area is

$A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2$

and the average current density is

$J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2$

8 0

2 years ago

Read 2 more answers

Which of the following statements accurately describes the atmospheric patterns that influence local weather?

timurjin [86]

Answer: A

Explanation:

Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.

4 0

1 year ago

The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie

Anni [7]

Answer:

The inverse frequency is $\dfrac{3}{80}\ s$

Explanation:

Given that,

Magnetic field = 20 T

Proportionality constant = 5 Hz/T

Change in magnetic field = 3 T

We know that,

$B=\dfrac{K}{\dfrac{1}{\omega}}$

We need to calculate the inverse frequency

Using formula of frequency

$\Delta(\dfrac{1}{\omega})=\dfrac{\Delta B}{k\times(\dfrac{1}{\omega^2})}$

$\Delta(\dfrac{1}{\omega})=\dfrac{k\times\Delta B}{B^2}$

Put the value into the formula

$\Delta(\dfrac{1}{\omega})=\dfrac{3\times5}{(20)^2}$

$\Delta(\dfrac{1}{\omega})=\dfrac{3}{80}\ s$

Hence, The inverse frequency is $\dfrac{3}{80}\ s$

5 0

2 years ago