×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
The molar volumes of ice and water are 0.0196 and 0.0180 litres per mole at 273K. If ΔH for the transition of ice to water is 1440 calories per mole at 1 atm pressure, find ΔU.

```
4 years ago

Vikas TU
14146 Points
```							From eqn.ΔU   +  Δ(PV)ΔH =         = ΔU   +  ΔPV + PΔV      Given,1440 = ΔU  + 0.0196*273 + 0.0180*273ΔU = 1440  – 10.2648      =  1429.73 calories.
```
4 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Physical Chemistry

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 141 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions