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 compressibility of a gas is less than unity at STP. Therefore,
a. Vm > 22.4 L
b. Vm < 22.4 L
c. Vm = 22.4 L
Vm = 44.8 L
```
5 years ago

396 Points
```							Sol. Compressibility factor (Z) = (V )/V_id  < 1 (given)
⇒ V < 22.4 L ∵ Vid (1 mol) = 22.4 L at STP

```
5 years ago
satyendra kumar
19 Points
```							z=molar volume of real gas/molar volume of ideal gas           *molar volume of ideal gas at STP is 22.4L*for z less than 1,molr volume of real gass  should be less than 22.4L
```
2 years ago
Hitanshu Sekhar
11 Points
```							Given that, Z ( compressibility factor ) of a gas We know that, Z = pV/nRTFor an ideal gas Z=1At high pressure value of Z increases because repulsion b/w molecules gives a noticable effect making molar volume Vm of the real gas greater than the molar volume of the ideal gas (Vm (ideal gas ) = RT/p ) causing Z to exceed 1. When pressures are lower, attractive forces dominate making ZSo, if we increase the molar volume Vm of a gas, the pressure decreases, as a result attractive forces dominate making Z So, the answer is Vm>22.4L.
```
2 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