Question

At what temperature will 41.6 g of N2(g) exert a pressure of 108.6 kPa in a 20.0L cylinder?

Question 7 options:

134 K


176 K


238 K


337 K

Answer 1

Answer:

238 K

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Answer 2

Step by step explanation:

Given :  $41.6 g$ of $N_2(g)$ exert a pressure of $108.6 kPa$ in a $20.0L$ cylinder.

To find : The temperature of cylinder.

Formula used : We use Ideal gas equation to find the temperature

$PV=nRT$

P is the pressure, V is the volume, n is the no. of moles, R is the universal gas constant and T is the temperature.

and to find temperature(T),

$T=\frac{PV}{nR}$

No. of moles(n) for any compound 'a' are,

$n_a=\frac{W_a}{MM_a}$

W is the given mass and MM is the molar mass of compound 'a'.

Given data :

$P=108.6kPa$

$V=20.0L$

Mass of $N_2(g)$ $= 41.6g$

• Required conversion units

Pressure : $1kPa=1000Pa$

                 $1Pa=\frac{1}{101325} atm$

• Required data for calculation of temperature

⇒  $P=108.6kPa$

        $=108.6*1000Pa$

        $=\frac{108.6*1000}{101325} atm$

        $=1.072atm$

⇒  $V=20.0L$

Mass of $N_2(g)$ $= 41.6g$

Molar mass of $N_2(g)$ $=28g/mol$

by using formula the no. of moles are,

$n_a=\frac{W_a}{MM_a}$

⇒  $n=\frac{41.6g}{28g/mol}$

        $=1.486mol$

⇒  $R=0.0821atmLK^{-1}mol^{-1}$

• Calculation for Temperature,

Now we have all required data in required units,

$P=1.072atm$  ,  $V=20.0L$  ,  $n=1.486mol$   and $R=0.0821atmLK^{-1}mol^{-1}$

by using Ideal gas equation,

⇒  $T=\frac{PV}{nR}$

        $=\frac{1.072atm*20.0L}{1.486mol*0.0821atmLK^{-1}mol^{-1}}$

        $=\frac{21.44}{0.122K^{-1}}$

        $=175.73K$

        $\approx 176K$

Hence we get the temperature of cylinder containing $N_2(g)$ is  (b) $176K$ .