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diamong [38]
2 years ago
12

A gas occupies 2 m^3 at 27°C at a pressure of 1 atmosphere. At a pressure of 2 atmospheres it occupies a volume of 1 m^2. What i

s its temperature at this new volume and pressure?
Physics
1 answer:
diamong [38]2 years ago
6 0

Answer:

27°C

Explanation:

We'll begin by converting 27 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 27 °C

Initial temperature (T₁) = 27 °C + 273

Initial temperature (T₁) = 300 K

Next, we shall determine the final temperature of the gas. This can be obtained as follow:

Initial volume (V₁) = 2 m³

Initial temperature (T₁) = 300 K

Initial pressure (P₁) = 1 atm

Final pressure (P₂) = 2 atm

Final volume (V₂) = 1 m³

Final temperature (T₂) =?

P₁V₁/T₁ = P₂V₂/T₂

1 × 2 / 300 = 2 × 1 / T₂

2/300 = 2/T₂

1/150 = 2/T₂

Cross multiply

T₂ = 150 × 2

T₂ = 300 K

Finally, we shall convert 300 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 300 K

T(°C) = 300 – 273

T(°C) = 27°C

Thus, the final temperature is 27°C

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It is night. Someone who is 4 feet tall is walking away from a street light at a rate of 8 feet per second. The street light is
KiRa [710]

Answer: 4 ft/s

Explanation:

Given

height of man=4 ft

speed of person v=8 ft/s

height if street light=12 ft

Let x be the distance between person and street light and y be the length of his shadow

From diagram

as the two triangle ADE and ABC are similar therefore we can say that

\frac{4}{12}=\frac{y}{x+y}

\frac{1}{3}=\frac{y}{x+y}

x+y=3y

x=2y

differentiate above Equation w.r.t time we get

\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{8}{2}=4 ft/s

6 0
2 years ago
the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is
garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

6 0
3 years ago
How would you change the distance between two charged particles to increase the electric force between them by a factor of 16
Naily [24]

The electrostatic force between two charges is inversely
proportional to the square of the distance between them.

So if you want to multiply the force by, say, ' Q ',
you need to multiply the distance by  ( 1 / √Q ) .

We want to multiply the force by 16, so we need to
multiply the distance by    ( 1 / √16 )  =  ( 1 / 4 ) .

The distance should be changed to  1/4  of what it is now.

4 0
3 years ago
Read 2 more answers
An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do
ipn [44]

Answer:

(a) ΔU=747J

(b) γ=1.3

Explanation:

For (a) change in internal energy

According to first law of thermodynamics the change in internal energy is given as

ΔU=Q-W

Substitute the given values

ΔU=970J-223J

ΔU=747J

For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

Cv is the molar heat capacity at constant volume

To calculate γ we first need to find Cp and Cv

So

For Cp

As we know

Q=nCpΔT

Cp=(Q/nΔT)

C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\

So

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

γ=1.3

4 0
3 years ago
Can someone please explain number 8?
guajiro [1.7K]

Answer: Ax=(Vx-Vox)/(T)

Vx=Vox+Ax*T

Solving for Ax in terms of Vx, Vox, T

Vx-Vox=Ax*t

Ax=(Vx-Vox)/(T)

This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.

Any questions please feel free to ask. Thanks

8 0
2 years ago
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