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bagirrra123 [75]
3 years ago
10

A 0.216 g sample of carbon dioxide, CO2, has a volume of 507 mL and a pressure of 470 mmHg. What is the temperature of the gas i

n kelvins?
Chemistry
1 answer:
Gala2k [10]3 years ago
6 0

Answer:

The temperature of the gas is 876.69 Kelvin

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

  • P= 470 mmHg
  • V= 570 mL= 0.570 L
  • n= 0.216 g= 0.0049 moles (being the molar mass of carbon dioxide is 44 g/mole)
  • R= 62.36367 \frac{mmHg*L}{mol*K}
  • T=?

Replacing:

470 mmHg*0.570 L= 0.0049 moles* 62.36367 \frac{mmHg*L}{mol*K} *T

Solving:

T=\frac{470 mmHg*0.570 L}{0.0049 moles* 62.36367\frac{mmHg*L}{mol*K} }

T= 876.69 K

<em><u>The temperature of the gas is 876.69 Kelvin</u></em>

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Explanation :

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and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g

Now we have to calculate the final temperature of the mixture.

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

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.42J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol = 35.5 g

m_2 = mass of water = 45.0 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 8.0^oC

T_2 = initial temperature of water = 28.6^oC

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

35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.14^oC

Therefore, the final temperature of the mixture is, 22.14^oC

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