Ask question

Ask question

All categories

4 years ago

8

During a compression at a constant pressure of 290 Pa, the volume of an ideal gas decreases from 0.62 m3 to 0.21 m3. The initial

temperature is 320 K, and the gas loses 160 J as heat. What are (a) the change in the internal energy of the gas and (b) the final temperature of the gas?

1 answer:

Aloiza [94]4 years ago

5 0

Answer:

a) -41.1 Joule

b) 108.38 Kelvin

Explanation:

Pressure = P = 290 Pa

Initial volume of gas = V₁ = 0.62 m³

Final volume of gas = V₂ = 0.21 m³

Initial temperature of gas = T₁ = 320 K

Heat loss = Q = -160 J

Work done = PΔV

⇒Work done = 290×(0.21-0.62)

⇒Work done = -118.9 J

a) Change in internal energy = Heat - Work

ΔU = -160 -(-118.9)

⇒ΔU = -41.1 J

∴ Change in internal energy is -41.1 J

b) V₁/V₂ = T₁/T₂

⇒T₂ = T₁V₂/V₁

⇒T₂ = 320×0.21/0.62

⇒T₂ = 108.38 K

∴ Final temperature of the gas is 108.38 Kelvin

You might be interested in

someone help me please

Lemur [1.5K]

Answer:

but for helping we need yr ques nd they r not here

4 0

3 years ago

Read 2 more answers

PLEASE HELP I HAVE FINALSSS

vitfil [10]

Answer: 26.67 m/s

Explanation:

Given

Length traveled by the ball $s=40\ m$

Time taken to reach the goal post is $t=3\ s$

Initial velocity $u=0\ m/s$

Using the second equation of motion

$\Rightarrow s=ut+\frac{1}{2}at^2\\\Rightarrow 40=0+\frac{1}{2}a(3)^2\\\Rightarrow a=\frac{80}{9}\ m/s^2\\$

Now using

$\Rightarrow v^2-u^2=2as\\\\\Rightarrow v^2-0=2\times \frac{80}{9}\times 40\\\\\Rightarrow v=\sqrt{\dfrac{80\times 80}{9}}=\dfrac{80}{3}\\\Rightarrow v=26.67\ m/s$

The velocity of ball will be 26.67 m/s

8 0

3 years ago

A certain sprinter has a top speed of 10.5 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able

Svet_ta [14]

Answer:

Total time = 10.667 seconds

Explanation:

For the first 12m

V² = u² + 2as

10.5² = 0² + 2a(12)

110.25 = 24a

a = 4.59375 m/s²

V = u + at

10.5 = 0 + 4.59375t

t = 10.5/4.59375

t = 2.286 seconds

For the remaining race

100-12 = 88m

He travels at a constant speed for 88m

S = ut + 1/2(at²)

But a = 0

S = ut

88= 10.5t

t = 88/10.5

t = 8.38.seconds

Total time = 2.286 + 8.381

Total time = 10.667 seconds

4 0

4 years ago

A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

Murrr4er [49]

Answer:

$\omega_2=5.1rad/s$

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is $L=I\omega$ . If we call 1 the situation when the student is at the rim and 2 the situation when the student is at $r_2=1.39m$ from the center, then we have:

$L_1=L_2$

Or:

$I_1\omega_1=I_2\omega_2$

And we want to calculate:

$\omega_2=\frac{I_1\omega_1}{I_2}$

The total moment of inertia will be the sum of the moment of intertia of the disk of mass $m_D=119.1 kg$ and radius $r_D=3.23m$ , which is $I_D=\frac{m_Dr_D^2}{2}$ , and the moment of intertia of the student of mass $m_S=54.3kg$ at position $r$ (which will be $r_1=r=3.23m$ or $r_2=1.39m$ ) will be $I_{S}=m_Sr_S^2$ , so we will have:

$\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}$

or:

$\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}$

which for our values is:

$\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s$

6 0

3 years ago

A 2mC charge traveling 10 m/s nears a wire carrying a 5 A current. If the charge's velocity and the wire's current are perpendic

-BARSIC- [3]

Answer:

The magnetic force on the wire at the moment is 2 micro-Newton/(Ampere-meter)

Explanation:

Formula for magnetic force is F = qvB*sin(theeta)

and B = μ*I / 2*pi*r

where

q = charge in coulomb

v = velocity

B = magnetic field strength

μ = permeability of free space

I = current

so from here we get B = (4*pi*10^(-7))(5) / 2*pi*0.01 = 0.0001-T

now ,

F = (2mC)*(10)*0.0001*sin(90)

F = 2 micro-Newton/(Ampere-meter)

3 0

3 years ago