Question

what is the temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm?

Answer 1

Answer:

—96.03°C

Explanation:

We'll begin by writing out the information provided by the question. This includes:

Number of mole (n) = 0.645 mole

Volume (V) = 2.00 L

Pressure (P) = 4.68 atm

Temperature (T) =?

Recall: that the gas constant = 0.082atm.L/Kmol

With the ideal gas equation PV = nRT, the temperature of the gas can be obtained as follow:

PV = nRT

4.68 x 2 = 0.645 x 0.082 x T

Divide both side 0.645 x 0.082

T = (4.68 x 2) /(0.645 x 0.082)

T = 176.97 K

Now, We can also express the temperature obtained in celsius as shown below:

Temperature (celsius) = temperature (Kelvin) - 273

Temperature (celsius) = 176.97 - 273

Temperature (celsius) = —96.03°C

The temperature of the Neon gas is

—96.03°C

Answer 2

The temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm is 176.6K

HOW TO CALCULATE TEMPERATURE:

• The temperature of an ideal gas can be calculated using the following formula:

PV = nRT

Where;

1. P = pressure (atm)
2. V = volume (L)
3. R = gas law constant (0.0821 molL/atmK)
4. T = temperature (K)

• According to this question, 0.645 mole of neon in a 2.00 L vessel is at 4.68 atm. The temperature can be calculated as follows:

4.68 × 2 = 0.645 × 0.0821 × T

9.36 = 0.053T

T = 9.36 ÷ 0.053

T = 176.6K

• Therefore, the temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm is 176.6K.

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