Question

At what temperature would 2.10 moles of Nitrogen gas have a pressure of 1.25 atm and in a 25.0 L tank? (Hint: R = 0.082 L atm K−1 mol−1)

Answer 1

Answer:

Given :-

• Moles = 2.10
• Gas = Nitrogen
• Pressure = 1.25 atm
• Volume = 25 l

To Find :-

Temperature

Solution :-

As we know that

PV = nRT

Here,

P = Pressure

V = Volume

n = Moles

T = Temperature

1.25 × 25 = 2.1 × 0.082 × T

31.25 = 2.1 × 0.082 × T

31.25 = 1.722T

31.25/0.1722 = T

181 ≈ T

Temperature is 181 K

$\huge\fbox{181 K {}}$

Answer 2

Given: Mole = 2.10

          Pressure = 1.25 atm

          Volume = 25L

          R = 0.082 L atm⁻¹mole⁻¹

To find: Temperature

Solution: According to the ideal gas law,

PV = nRT

Where P = pressure, V = volume of the gas, n = number of moles, R = universal gas constant, T is the temperature of the gas.

1.24 × 25 = 2.10 × 0.082 × T

T = 1.24×25/0.082

  = 43K or -230 °C

Therefore, the temperature of the 2.10 moles of Nitrogen gas having a pressure of 1.25 atm and in a tank of 25L is 43K.

• The ideal gas equation is the gas equation of a hypothetical gas that obeys all the hypotheses of the gas.
• The ideal gas equation gives the effect of pressure and temperature on the volume of a gas is called the ideal gas equation.
• It is also known as combined gas law.
• PV = nRT is the equation of ideal gas law.