For a liquid- the normal boiling point is -173- C- then at 2 atm pressure its boiling point should be nearly - H vap -200 cal - mole - R - -2 cal mol -1 K -1- In 2-0-7A- -73- CB- -333- CC- -60- CD- -103- C

The correct option is C 60^∘ CT_1 = 173^∘ C=100k P_1 = 1atm P_2 = 2atm Δ H_vap = 200 cal/mole P_2/P_1 = Δ H_vap/2 (T_2 T_1/T_1 T_2) In (2/1) = (200/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Friction

For a liquid,...

Question

For a liquid, the normal boiling point is $- 173^{\circ} C$ , then at 2atm pressure its boiling point should be nearly $(Δ H_{v a p} = \frac{200 c a l}{m o l e}, R = 2 c a l m o l^{- 1} K^{- 1}, I n 2 = 0.7)$

- 73^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 333^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 60^{\circ} C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 103^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 60^{\circ} C$
$T_{1} = - 173^{\circ} C = 100 k P_{1} = 1 a t m P_{2} = 2 a t m Δ H_{v a p} = 200 c a l / m o l e \frac{P_{2}}{P_{1}} = \frac{Δ H_{v a p}}{2} (\frac{T_{2} - T_{1}}{T_{1} T_{2}}) I n (\frac{2}{1}) = (\frac{200}{2}) (\frac{1}{100} - \frac{1}{T_{2}}) T_{2} = 333 K = 60^{\circ} C$

Suggest Corrections

Similar questions

Q. The normal boiling point of a liquid A is 350 K.

Δ H_{v a p}

at normal boiling point is 35 KJ/mole. Pick out the correct statement(s). (Assume

Δ H_{v a p}

to be independent of pressure).

Q. If the boiling point of water in Kelvin scale is 373 K. What is the boiling point of water on degree celcilus scale(

^{o}

C)?

The boiling point of liquid argon is :

$(a) - 196^{\circ} C (b) - 183^{\circ} C (c) - 186^{\circ} C (d) - 193^{\circ} C$

Q. The boiling point of a liquid is 50 K at 1 atm and

Δ H_{v a p} = 460.6

cal/mol. What will be its boiling point at 10 atm pressure ?

Q. For a sucrose solution, elevation in boiling point is

{1.0}^{\circ} C

, then what will be the boiling point of NaCl solution for same molal concentration?

Join BYJU'S Learning Program

For a liquid, the normal boiling point is −173∘C, then at 2atm pressure its boiling point should be nearly (ΔHvap=200calmole,R=2cal mol−1K−1,In2=0.7)

The correct option is C −60∘CT1=−173∘C=100kP1=1atmP2=2atmΔHvap=200cal/moleP2P1=ΔHvap2(T2−T1T1T2)In(21)=(2002)(1100−1T2)T2=333K=60∘C

For a liquid, the normal boiling point is $- 173^{\circ} C$ , then at 2atm pressure its boiling point should be nearly $(Δ H_{v a p} = \frac{200 c a l}{m o l e}, R = 2 c a l m o l^{- 1} K^{- 1}, I n 2 = 0.7)$