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2 years ago

A tank is filled with an ideal gas at 400 K and pressure of 1.00 atm.

Physics

1 answer:

bekas [8.4K]2 years ago

6 0

To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.

Mathematically it can be defined as

$PV=nRT$

Where

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

When the number of moles and volume is constant then the expression can be written as

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or in practical terms for this exercise depending on the final temperature:

$T_2 = \frac{P_2T_1}{P_1}$

Our values are given as

$T_1 = 400K\\P_1 = 1atm\\P_2 = 2atm$

Replacing

$T_2 = \frac{(2)(400)}{1}\\T_2 = 800K$

Therefore the final temperature of the gas is 800K

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A mass of 25.5 g of H2O(g) at 373 K is mixed with 325 g of H2O(l) at 285 K and 1 atm. Calculate the final temperature of the sys

Vedmedyk [2.9K]

Answer:

331.28 K

Explanation:

To solve this problem, you need to know that the heat that the water at 373 K is equal to the heat that the water at 285 K gains.

First, we will asume that at the end of this process there won't be any water left in gaseous state.

The heat that the steam (H20(g)) loses is equal to the heat lost because the change of phase plus the heat lost because of the decrease in temperature:

$Q_g = c_l*m_{wg} + m_{wg}*c*(T_{og}-T{fg})$

The specific Heat c of water at 298K is 4.18 kJ/K*kg.

The latent heat cl of water is equal to 2257 kJ/kg.

The heat that the cold water gains is equal to heat necessary to increase its temperature to its final value:

$Q_l = m_{wl}*c*(T_{fl}-T_{ol})$

Remember that in equilibrium, the final temperature of both bodies of water will be equal.

Then:

$Q_g = Q_l\\c_l*m_{wg} + m_{wg}*c*(T_{og}-T{fg}) = m_{wl}*c*(T_{fl}-T_{ol})\\2257 kJ/kg*0.0255 kg + 0.0255 kg*4.18 kJ/kg*K*(373K - T_f) = 0.325 kg*4.18kJ/kg*K*(T_f-285K)\\57.5535 kJ + 39.75807kJ - 0.10659T_f = 1.3585 T_f - 387.1725 kJ\\484.48407 kJ = 1.46244 T_f\\T_f = 484.48407 kJ /1.46244 = 331.28 K$

4 0

3 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

2 years ago

A man wears convex lens glasses of focal length 30cm in order to correct his eyes defect. Instead of the optimum 25cm, his dista

omeli [17]

Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

30 = (frac {25 {times} (-v)}{25 – (-v)})

v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

7 0

2 years ago

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves

KIM [24]

Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

Explanation:

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

4 0

3 years ago

An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit

lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is $1.127\times 10^7m/s$