A flask is filled with 13 gm of an ideal gas at 27°C and its temperature is raised to 52°C. The mass of the gas that has to be released to maintain the temperature of the gas in the flask at 52°C and the pressure remaining the same is

The correct option is 1.0 g.

To find the mass of the gas that needs to be released from the flask, we can apply the ideal gas equation:
$PV=nRT$

We know that the number of moles ($n$) is given by the mass of the gas ($m$) divided by its molar mass ($M$). Thus, the equation can be rewritten as:
$PV=\frac{m}{M}RT$

Since the volume of the flask ($V$) remains constant, the pressure ($P$) is maintained the same, and the gas constant ($R$) and molar mass of the gas ($M$) are constants, the product of the mass of the gas and its absolute temperature must be constant:
$mT=\text{constant}$
This gives us the relation:
$m_1{T}_1=m_2{T}_2$

First, convert the temperatures from Celsius to Kelvin:
Initial temperature, $T_1=27+273=300\text{K}$
Final temperature, $T_2=52+273=325\text{K}$

Given the initial mass of the gas is $m_1=13\text{g}$, we can solve for the final mass ($m_2$) of the gas that remains in the flask at $52°\text{C}$:
$13\times 300=m_2\times 325$

$m_2=\frac{13\times 300}{325}$

$m_2=\frac{3900}{325}=12\text{g}$

The mass of the gas that has to be released is the difference between the initial mass and the remaining mass:
$\text{Mass released}=m_1-m_2$

$\text{Mass released}=13\text{g}-12\text{g}=1.0\text{g}$