Helium gas with a volume of3.20 L, under a pressure of 0.180 atmand at$\mathbf{41}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$, is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 g/mol.

The ideal gas equation relates the variables pressure P, volume V, temperature T, and number of moles n as PV-nRT …….(1) where R is the proportionality constant called the Universal gas constant.

In this question, the number of moles is kept constant. Then the above expression can be rewritten as$\frac{PV}{T}=nR$, which is thus,$\frac{PV}{T}$=constant

For two distinct cases of each variable, the ideal gas equation can be written as .

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

The data given includes initial pressure , initial volume , $P_1$initial temperature , final pressure , final volume $V_1$. Rearranging the above equation for finding , as $T_2=\frac{P_2{V}_2{T}_1}{P_1{V}_1}$…………………(2)

The variable is the number of moles which is expressed as$n=\frac{n}{M}$ where is the mass of the sample taken, and is the molar mass of the compound. This gives the expression for finding the mass as $m=n*M$…………………(3)

The temperature after warming is calculated by substituting the data given in the equation of (2).

Given data are $P_1=0.180atm,T_1=41.0^{0}C(=314K),V_1=3.20L(=3.20*{10}^{-3}{m}^3)$, the new pressure and temperature after warming are double the initial value. Thus,$P_2=0.36atm,V_2=6.40L(=6.40*{10}^{-3}{m}^3),T_2=?$

The new temperature is

$$\begin{gathered} T_2=\frac{0.360*6.40*{10}^{-3}*314}{0.180*3.20*{10}^{-3}} \\ {T}_2=1256K \end{gathered}$$

Hence, the final temperature is 1256K, which is four times the initial temperature.

To find the mass, $m=n*M$, the number of moles is found from the ideal gas equation as $PV=nRTasn=n=\frac{PV}{RT}$.

Take the values of $P_1=0.180atm,T_1=41.0^{0}C(=314K),V_1=3.20L,R=0.08205746Latm{K}^{-1}mo{l}^{-1}$thus,$n=\frac{P_1{V}_1}{R{T}_1}$ .

Number of moles is found as

$$\begin{gathered} n=\frac{0.180*3.20}{0.08205746*314} \\ n=0.0223moles \end{gathered}$$

Thus the number of moles of He is $0.0223moles$.

Substituting the values of molar mass M = 4.00g / mol, and the number of moles $n=0.0223moles$in $m=n*M$.

$$\begin{gathered} m=0.0223*4.00 \\ m=0.0892g \end{gathered}$$

Mass of He taken is $m=0.0892g$.