You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

1st Law of Thermodynamics

1 g of H2O ...

Question

$1 g$ of $H_{2} O$ changes from liquid to vapour phase at constant pressure at $1 a t m$ . The volume increases from $1 c c t o 1671 c c$ . The heat of vaporisation at this pressur is $540 c a l / g$ . The increase in internal energy of water is

2099 J

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

3000 J

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

992 J

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

2122 J

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

Open in App

Solution

The correct option is A $2099 J$
Work done is given as-
$W = P (d V) = 1.01 \times 10^{5} (1671 - 1) \times 10^{- 6} = 167 J$
$Δ Q$ = $Δ U$ + $Δ W$
or
$Δ U$ = $Δ Q$ - $Δ W$
$= m L - 167 = (540 \times 4.2) - 167 = 2099 J$

Suggest Corrections

Similar questions

Q. When

1 g

of water changes from liquid to vapour phase at constant pressure of

1 a t m

, the volume increases from

1 c c

1671 c c

. The heat of vaporization at the pressure is

540 c a l / g

. Find the increase in internal energy of water.

Q. When

1 g m

of water changes from liquid to vapour phase at constant pressure of

1

atmosphere, the volume increases from

1 c c

1671 c c

. The heat of vaporisation at this pressure is

540 c a l / g m

. Increase in internal energy of water is

(1 a t m o s p h e r e = 1.01 \times 10^{6} d y n e / c m^{2})

Q. When 1gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increase from 1

c m^{3}

to 1671

c m^{3}

. The heat of vaporisation at this pressure is 540 cal/gm. Find (a) the work done (in J) in change of phase and (b) increase in internal energy of water.(1 atmosphere =

1.01 \times 10^{6} d y n e / c m^{2})

Q. 1 g of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increases from 1 mL to 1671 mL. The heat of vapourisation at this pressure is 540cal/g.The work done (in J) during phase change is

The latent heat of Vaporisation of a liquid at 500K and 1atm pressure is 30 Kcal/mole. What will be the change in internal energy of 3 mol of liquid at same temperature?

Join BYJU'S Learning Program

1 g of H2O changes from liquid to vapour phase at constant pressure at 1 atm. The volume increases from 1 cc to 1671 cc. The heat of vaporisation at this pressur is 540 cal/g. The increase in internal energy of water is

The correct option is A 2099JWork done is given as-W=P(dV)=1.01×105(1671−1)×10−6=167JΔQ = ΔU + ΔWor ΔU = ΔQ - ΔW=mL−167=(540×4.2)−167=2099J

$1 g$ of $H_{2} O$ changes from liquid to vapour phase at constant pressure at $1 a t m$ . The volume increases from $1 c c t o 1671 c c$ . The heat of vaporisation at this pressur is $540 c a l / g$ . The increase in internal energy of water is

The correct option is A $2099 J$
Work done is given as-
$W = P (d V) = 1.01 \times 10^{5} (1671 - 1) \times 10^{- 6} = 167 J$
$Δ Q$ = $Δ U$ + $Δ W$
or
$Δ U$ = $Δ Q$ - $Δ W$
$= m L - 167 = (540 \times 4.2) - 167 = 2099 J$