Question

An open vessel containing air at 292 k was cooled to a certain temperature at which the number of moles of the gaseous molecules increased by 25% the final temperature of the vessel is

Answer 1

Answer:

The final temperature of the vessel would be 233.6K.

Explanation:

We can readily solve this question with the help of the ideal gas equation.

The mathematical form of the ideal gas equation is:

PV=nRT

⇒ $\frac{PV}{nRT} = constant$

Thus, we can say that,

$\frac{P_{1} V_{1} }{n_{1} R T_{1} } = \frac{P_{2} V_{2} }{n_{2} R T_{2} }= constant$

According to the given information, in this case, $P_{1} , V_{1} ; P_{2} ,V_{2}$ remain unchanged since the vessel is open. [R= universal gas constant= constant]

Thus, the base form of the equation we will use here to solve the problem is:

⇒ $\frac{n_{1} }{n_{2} } = \frac{T_{2} }{T_{1} }$

Calculation:

Now, let us assume the number of moles present initially $(n_{1} )$ is 100x.

∴ As per the question, the final number of moles increased was by 25%

    ⇒ $n_{2} = 125x$

∴ By the problem,

⇒ $\frac{n_{1} }{n_{2} } = \frac{T_{2} }{T_{1} }$

⇒ $T_{2}= \frac{n_{1} }{n_{2} } * T_{1}$

⇒ $T_{2} = \frac{100x}{125x} * 292 = 233.6K$

#SPJ2