Question

Calculate the temperature at which 28 g of N2 will occupy a volume of 10.0 liters at 2.46
atmospheric pressure.

Answer 1

Through gas equation Pv=nrt 2.46*10=28/28*0.0821*t 5604k

Answer 2

Answer: The temperature of nitrogen gas is 300 K.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of nitrogen gas = 28 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$\text{Moles of nitrogen gas}=\frac{28g}{28g/mol}=1mol$

To calculate the temperature of the gas, we use the equation given by ideal gas, which is:

$PV=nRT$

P = pressure = 2.46 atm

V = volume = 10 L

n = number of moles = 1 mol

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature = ?

Putting values in above equation, we get:

$2.46atm\times 10L=1\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times T\\\\T=300K$

Hence, the temperature of the nitrogen gas is 300 K.