Question

A gas in an open container is heated from ${27}^{0}C$ to $1{27}^{0}C$. The fraction of the original amount of the gas escaped in the container will be:

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{3}{4}$
• D. none of these

Answer 1

The correct option is C $\frac{3}{4}$

PV = nRT = constant

Here Pressure P and Volume V are constant in this case,

n is number of moles

R is gas constant

T is temperature

Given temperatures are 27 and 127 degree celcius

$T_1$ = 27 + 273 = 300 K

$T_2$ = 127 + 273 = 400 K

So,

$n_1$ $T_1$ = $n_2$ $T_2$

$n_1\times 300$ = $n_2\times 400$

$n_2$= $\frac{3n_1}{4}$

So the fraction of the original amount of the gas escaped is 3/4