Physics, 21.02.2020 02:28 pg67891012345

A sample of nitrogen occupies 5.50 liters under a pressure of 900 torr at 25oC. At what temperature will it occupy 10.0 liters at the same pressure?

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Answer from: TightKnowsDaWhey

The temperature at which the gas will occupy 10 L is 268.8 °C

We'll begin by listing out what was given from the question. This includes:

Initial volume (V₁) = 5.5 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

Final volume (V₂) = 10 L

Pressure = constant

Final temperature (T₂) =?

Using the Charles' law equation, we can obtain the new temperature of the gas as illustrated below:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} \\\\\frac{5.5}{298} = \frac{10}{T_{2}} \\\\$

Cross multiply

5.5 × T₂ = 298 × 10

5.5 × T₂ = 2980

Divide both side by 5.5

$T_{2} = \frac{2980}{5.5}\\\\$

T₂ = 541.8

Subtract 273 to express in degree celsius (°C)

T₂ = 541.8 – 273

T₂ = 268.8 °C

Therefore, the temperature at which the gas will occupy 10 L is 268.8 °C

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Answer from: rachellynn02

At 268.82°C volume occupied by nitrogen is 10 liters at pressure of 900 torr.

Explanation:

Given:

Volume of a sample of nitrogen = 5.50 liters

Pressure = 900 torr

Temperature = 25°C

To find the temperature at which the nitrogen will occupy 10 liters volume at same pressure.

Solution:

Since the pressure is kept constant, so we can apply the temperature-volume law also called the Charles Law.

Charles Law states that the volume of a gas held at constant pressure is directly proportional to the temperature of the gas in Kelvin.

Thus, we have :

∝

$\frac{V}{T}=k$

where is a constant.

For two samples of gases, the law can be given as:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

From the data given:

$V_1=5.5\ l$

$T_1=25\ \°C =(273+25)K= 298 K$

$V_2=10\ l$

We need to find T_2 .

Plugging in values in the formula.

$\frac{5.5}{298}=\frac{10}{T_2}$

Multiplying both sides by T_2 .

$T_2\times\frac{5.5}{298}=\frac{10}{T_2}\times T_2$

$\frac{5.5}{298}T_2={10}$

Multiplying both sides by $\frac{298}{5.5}$

$\frac{298}{5.5}\times\frac{5.5}{298}T_2=\frac{10\times 298}{5.5}$

$T_2=541.82\ K$

$T_2=541.82\ K-273\ K = 268.82\°C$

Thus, at 268.82°C volume occupied by nitrogen is 10 liters at pressure of 900 torr.

Answer from: Quest

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Answer from: Quest

The relationship of both of them

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A sample of nitrogen occupies 5.50 liters under a pressure of 900 torr at 25oC. At what temperature...

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