Question

1g of helium gas is confined in 2 litr flask under a pressure of 2.05atm what is its temperature

Answer 1

HELLO FRIENDS
To solve use the following form of the ideal gas equation, T =  PV/nR 

P = 2.05 atm   V = 2L  n = 1g /4  g/mol = 0.25 mol  R = 0.0821 L atmK-1mol-1

​T = 2,05 x 2/ 0.25 x 0.0821 = 199.76 K

Answer 2

Answer: The temperature of the flask is 200 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 helium gas = 1 g

Molar mass of helium gas = 4 g/mol

Putting values in above equation, we get:

$\text{Moles of helium}=\frac{1g}{4g/mol}=0.25mol$

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

$PV=nRT$

P = pressure = 2.05 atm

V = volume = 2 L

n = number of moles = 0.25 mol

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

T = temperature = ?

Putting values in above equation, we get:

$2.05atm\times 2L=0.25\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times T\\\\T=200K$

Hence, the temperature of the flask is 200 K.