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For same mass of two different ideal gas...


For same mass of two different ideal gases of molecular weights `M_(1)` and `M_(2)` plots oflog V vs lot P at a given constant temperature are shown. Identify the correct option

A

`M_(1)gtM_(2)`

B

`M_(1)=M_(2)`

C

`M_(1)ltM_(2)`

D

Can be predicted only if temperature is known.

Text Solution

Verified by Experts

For ideal gases
`PV=nRT=w/MRT " " ( :. n=w//M)`
For same mass of gases and at constant temperature wRT is constant (K).
`:.PV=K/M`
Taking log both sides we get
`log P+logV="log"K/M`
`logV=-logP+"log"K/M`
comparing this with straight line equation
`y=mx+c`
For ideal gas 1, `("Intercept")_(1)="log"K/(M_(1))`
for ideal gas 2, `("Intercept")_(2)="log"K/(M_(2))`
`:. "log"K/(M_(2))gt"log"K/(M_(1))` (from graph)
or `K/(M_(2))gtK/(M_(1))impliesM_(1)gtM_(2)`
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