#### Thank you for registering.

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

Please check your email for login details.

Click to Chat

1800-1023-196

+91 7353221155

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

# Calculate the most probable velocity of CO2 molecules at 27 degree celcius

Grade:12th pass

## 2 Answers

Arun
25763 Points
3 years ago

molar mass

In case of oxygen gas, T = 27 + 273 = 300 K. R = 8.314 (kg m2 /s2 )/ K mol

Since molar mass of oxygen gas O2 = 32 gm/mol = 3.2 x10-2 Kg/mol

Hence,  μ = (3RT/M)1/2 = [3 x8.314 x300 / 3.2 x10-2 ]1/2 = [233831.25 ]1/2 =483.56 m/s

Hence, the r.m. speed fro oxygen gas at O0C (273K) = 483.5 m/s

Rajat Gaikar
13 Points
one year ago
For CO molar mass =12+16*2=44 g/mol=44*10-3 kg/mol
Vmp=(2RT/M)1/2=((2*8.314*300)/(44*10-3))1/2=336.70 m/s

## ASK QUESTION

Get your questions answered by the expert for free