365

You visited us 365 times! Enjoying our articles? Unlock Full Access!

Elevation in Boiling Point

Find the temp...

Question

Find the temperature at which $3$ moles of $S O_{2}$ will occupy a volume of $10$ litre at a pressure of $15 a t m$ .
$a = 6.71 a t m l i t r e^{2} m o l^{- 2}; b = 0.0564 l i t r e m o l^{- 1}$ .

Open in App

Solution

$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$
$(15 + \frac{6.71 \times 9}{100}) (10 - 3 \times 0.0564) = 3 \times 0.082 \times T$
$T = 624 K$

Suggest Corrections

Similar questions

S O_{2}

10

15 a t m

624 K

(a = 6.71 a t m L^{2} m o l^{- 2}; b = 0.0564 l i t r e m o l^{- 1})

in K

S O_{2}

10 L

15 atm

a = 6.71 {atm litre}^{2} {mol}^{- 2}

b = 0.0564 {litre mol}^{- 1}

in K

S O_{2}

10 L

15 atm

a = 6.71 {atm litre}^{2} {mol}^{- 2}

b = 0.0564 {litre mol}^{- 1}

192 g

S O_{2}

6 d m^{3}

15 a t m

[a = 5.68 a t m L^{2} m o l^{- 2}, b = 0.06 L m o l^{- 1}]

0^{o} C

Join BYJU'S Learning Program