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Brut [27]
2 years ago
14

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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3. What is the potential energy of a 8 Newton book sitting on a shelf that is 12 meters high?​
Svetradugi [14.3K]

Answer:

P = 96 J

Explanation:

Given that,

Weight of the book, W = mg = 8 N

It is placed at a height of 12 m

We need to find the potential energy of the book. The potential energy of an object is given by the formula as follows :

E = mgh

mg = Weight

E=8\ N\times 12\ m\\E=96\ J

So, the potential energy of the book is 96 J.

8 0
3 years ago
HELP ASAP PLZ
attashe74 [19]

Answer:

Increases

Explanation:

Because acceleration goes higher

5 0
3 years ago
A car is moving at 30.0 km/h when it accelerates at 2.0 m/s for 3.6 s. what is the car final speed?
Lana71 [14]

Answer:

wrong question

Explanation:

unit of acceleration is written in m/s which is wrong

3 0
3 years ago
Determine the electrical force of attraction between two balloons
Deffense [45]
Force = 9x10^9 x (6x10^-7 x 6x 10^-7) / (.5)^2
= 0.013 N

Using the formula;

F = 9x10^9 x (Q1 x Q2) / r^2
8 0
2 years ago
A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal
Natalija [7]

Answer : The specific heat of unknown sample is, 8748.78J/kg^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-[q_2+q_3]

m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]

where,

c_1 = specific heat of unknown sample = ?

c_2 = specific heat of water = 4186J/kg^oC

c_3 = specific heat of copper = 390J/kg^oC

m_1 = mass of unknown sample = 72.0 g  = 0.072 kg

m_2 = mass of water = 203 g  = 0.203 kg

m_2 = mass of copper = 187 g  = 0.187 kg

T_f = final temperature of calorimeter = 39.4^oC

T_1 = initial temperature of unknown sample = 80.0^oC

T_2 = initial temperature of water and copper = 11.0^oC

Now put all the given values in the above formula, we get

0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]

c_1=8748.78J/kg^oC

Therefore, the specific heat of unknown sample is, 8748.78J/kg^oC

7 0
3 years ago
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