The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen, and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is:

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{3 R T}{V}$

$\frac{4 R T}{V}$

$\frac{5}{2} \frac{R T}{V}$

$\frac{88 R T}{V}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} M o l e s o f o x y g e n = \frac{16}{32} = \frac{1}{2} \\ M o l e s o f n i t r o g e n = \frac{28}{28} = 1 \\ M o l e s o f c a r b o n d i o x i d e = \frac{44}{44} = 1 \\ T o t a l p r e s s u r e = S u m o f p a r t i a l p r e s s u r e s \\ {P=P}_{O}_{_{2}} {+P}_{N}_{_{2}} {+P}_{CO}_{2} \\ = \frac{1}{2} \frac{RT}{V} + \frac{RT}{V} + \frac{RT}{V} = \frac{5}{2} \frac{RT}{V} \end{array}$