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?

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Answer 1

Given :-

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

To Find :-

Temperature

Solution :-

As we know that

PV = nRT

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.

$\huge\fbox\red{181 }$

Answer 2

The required temperature is 181 K.

Number of moles of nitrogen gas$= 2.10\ moles$

Capacity of tank$=25.0 \ L$

The pressure of the tank$=1.25\ atm$

We need to find at what temperature, the given conditions will exist.

As we know,

$PV=nRT$

Where P is pressure,

V is volume,

n is the number of the moles,

R is the universal gas constant,

T is the temperature.

So,

$T=\frac{PV}{nR}$

$R=0.821\frac{L-atm}{mol}$

On putting the values in the formula, we get,

$T=\frac{1.25\times 25.0}{2.10\times 0.0821}$

$\implies T=181.25$

Hence the required temperature is 181 K.

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