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love history [14]
3 years ago
5

Two moles of neon gas at 25oC and 2.0 atm is expanded to 3 times the original volume while the pressure is reduced to 1.0 atm. F

ind the end temperature.
A. 447 C
B. 174 C
C. -66 C
D. 38 C
E. 150 C
Physics
1 answer:
bazaltina [42]3 years ago
5 0

Answer:

The end temperature is 174 °C

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.

So, being:

  • P= 2 atm
  • V=?
  • n= 2 moles
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 25 °C= 298 °K

and replacing:

2 atm*V= 2 moles* 0.082 \frac{atm*L}{mol*K} *298 K

you get:

V=\frac{2 moles* 0.082\frac{atm*L}{mol*K}  *298 K}{2 atm}

V= 24.436 L

Now, two moles of neon gas is expanded to 3 times the original volume while the pressure is reduced to 1.0 atm. Then you know:

  • P= 1 atm
  • V= 3*24.436 L=73.308 L
  • n= 2 moles
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= ?

Replacing:

1 atm*73.308 L= 2 moles* 0.082 \frac{atm*L}{mol*K} *T

Solving:

T=\frac{1 atm*73.308 L}{2 moles* 0.082\frac{atm*L}{mol*K}}

T= 447 °K= 174 °C (being 0°C=273 °K)

<u><em>The end temperature is 174 °C</em></u>

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Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

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