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Nostrana [21]
3 years ago
7

A lab technician has 45L gas in a flexible container at a pressure of 100.0 kPa. If the technician decreases the pressure to 60.

0 kPa, without changing the temperature, what is the new volume?
Chemistry
1 answer:
ICE Princess25 [194]3 years ago
5 0

Answer:

75 L

Explanation:

We assume that the gas in the container is an ideal gas, which is governed by the ideal gas equation : PV = nRT , where P is the pressure of the gas, V is the volume of the container, n is the number of mols of the gas, R is the universal gas constant and T is the temperature of the gas.

Here, R is a universal constant with a value of ≈ 8.314 J/mol-K. Also, given that temperature (T) is maintained constant. And since the container is closed, there is no exchange of the gas between the container and the surroundings, thus keeping the number of mols (n) of the gas constant. Hence, in the ideal gas equation, right hand side, i.e, nRT is a constant.

Initially the equation is: P_{1}V_{1} = nRT(constant)

Finally, the equation is:P_{2}V_{2} = nRT(constant)

Since, nRT is a constant, P_{1}V_{1} = P_{2}V_{2}

It is given that, P_{1} = 100.0KPa , V_{1} = 45L and P_{2} = 60.0KPa.

Therefore, 100x45 = 60 x V_{2} ⇒ V_{2} = 75L

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A sample of gas in a closed container at a temperature of 100oC and 3 atm is heated to 300oC. What is the pressure of the gas at
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6.14 atm

Explanation:

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Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

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Finally, we shall determine the pressure at the highest temperature as follow:

Initial temperature (T₁) = 373 K

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Final temperature (T₂) = 573 K

Final pressure (P₂) =?

P₁ / T₁ = P₂ / T₂

4 / 373 = P₂ / 573

Cross multiply

373 × P₂ = 4 × 573

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P₂ = 2292 / 373

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