Question

A monatomic ideal gas in a rigid container is heated from 206 K to 335 K by adding 8280 J of heat. How many moles of gas are there in the container?

Answer 1

Answer: 5.14 moles

Explanation:

heat supplied:  8280 J

The formula used for heat absorbed by an ideal gas:

$q=n\times C_v\times (T_2-T_1)$

where,

q = heat absorbed = 8280 J

$C_v$ = for monoatomic gas = $\frac{3}{2}R$

n = number of moles

R = gas constant = 8.314 J/Kmol

$T_1$ = initial temperature = 206 K

$T_2$ = final temperature  = 335 K

Now put all the given values in the above formula,

$8280J=n\times \frac{3}{2}\times (8.314J/Kmol)\times (335-206)K$

$n=5.14$

Thus there are 5.14 moles of gas in the container.