91Ó°ÊÓ

We need to find the grams of helium which added to the balloon.

For calculating the mass of the helium, we will use the ideal gas law equation

$PV=nRT...\left(1\right)$

Since, $n=\frac{mass}{molarmass}$

Substituting the value of $n$in equation $\left(1\right)$

role="math" localid="1652604534551" $PV=\frac{mass}{molarmass}RT...\left(2\right)$

Considering the given conditions at STP

The pressure $P=1atm$,

the temperature role="math" localid="1652606249977" $T=273K$,

the volume of gas $V=25L$

the universal gas constant $R=0.0821\frac{atm-L}{mol-K}$

the molar mass of helium $=4.003\frac{g}{mol}$

Now, for the mass of the Helium using the equation $\left(2\right)$

$mass=\frac{PV×molarmass}{RT}$

Substituting the given values

role="math" localid="1652606320437" $$\begin{gathered} mass=\frac{1atm×25L×4.003{\displaystyle \frac{g}{mol}}}{0.0821{\displaystyle \frac{atm-L}{mol-K}}×273K}=\frac{100.075}{22.41}g \\ mass=4.47g \end{gathered}$$

We need to find the final pressure.

For calculating the final pressure of Helium inside the balloon, we will use the combined gas law equation

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}...\left(1\right)$

Considering the given values

initial pressure role="math" localid="1652606723443" $P_1=1atm=760mmHg$,

initial volume $V_1=25L$

initial temperature $T_1=273K$, and

the final volume role="math" localid="1652606435110" $V_2=2460L$,

the final temperature $T_2=-35°C=\left(-35+273\right)K=238K$

Now, calculating the final pressure using the equation $\left(1\right)$

$P_2=\frac{P_1{V}_1{T}_2}{V_2{T}_1}$

Substituting the given values

role="math" localid="1652606945347" $$\begin{gathered} P_2=\frac{760mmHg×25L×238K}{2460L×273K}=\frac{4,522,000}{371,580}mmHg \\ {P}_2=6.73mmHg \end{gathered}$$