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2 years ago

What is the volume in (a) liters and (b) cubic yards of a room that is 10. meters wide by 15 meters long and 8.0 ft high?

Chemistry

1 answer:

deff fn [24]2 years ago

7 0

Answer:

V = 364500 L, 476.748 yard³

Explanation:

Given that,

The dimensions of a room are 10 meters wide by 15 meters long and 8.0 ft high.

l = 10 m, b = 15 m, h = 8 ft = 2.43 m

The volume of the room is :

V = lbh

So,

V = 10×15×2.43

V = 364.5 m³

As 1 m³ = 1000 L

364.5 m³ = 364500 L

Also, 1 m³ = 1.30795 yard³

364.5 m³ = 476.748 yard³

Hence, this is the required solution.

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b) 1200 K

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According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 6.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas = 1

$V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L$

a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

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

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?$

Putting values in above equation, we get:

$\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K$

The final temperature is 900 K

b) The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 6.00 atm

$P_2$ = final pressure of gas = $2\times 6.00atm=12.0atm$

$V_1$ = initial volume of gas = 4.10 L

$V_2$ = final volume of gas = $2\times 4.10 L=8.20L$

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas =?

Now put all the given values in the above equation, we get:

$\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}$

$T_2=1200K$

The final temperature is 1200 K

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3 years ago