Question

At what temperature the volume of 28 g of N 2 gas of
2.56 atm pressure will be 10 L ?? ​

Answer 1

Answer:

411.8K approx

Explanation:

P=2.56 atm

V=10L

n=28/28=1

R=0.0821 atmL /Kmol

applying PV=nRT

T=411.8 K approximately

Answer 2

Answer:

Given Mass = 28g

Molar Mass of Nitrogen = 28g

Therefore, Number of moles of Nitrogen is:

⇒ n = 28/28 = 1 mole.

According to the ideal gas equation,

⇒ PV = nRT

'P' is the pressure, 'V' is the volume, 'n' is the number of moles, 'R' is the universal gas constant and 'T' is the temperature.

Given Data:

• P = 2.56 atm
• V = 10 L
• n = 1 mole
• T = ?

Since, we have Pressure in atm and Volume in Liters, we would consider 'R' to be: 0.082 atm.L/K.mole

Substituting all the known information we get:

$\implies PV = nRT\\\\\implies 2.56 \times 10 = 1 \times 0.082 \times T\\\\\implies T = \dfrac{2.56 \times 10}{1 \times 0.082}\\\\\\\implies T = \dfrac{25.6}{0.082}\\\\\\\implies \boxed{ \bf{ T = 312.19 \; K}}$

This is the required temperature.