During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temperature T and volume V, when it expands to volume 2 V, the resulting temperature is

PV=RT= constant P=RTV VP2= constant ⇒ V×R2T2V2= constant R2T2V= constant ⇒ T2V= Constant T2V= T'22V or T'2=2T2 T' = 2T

During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temperature T and volume V, when it expands to volume 2 V, the resulting temperature is

PV=RT= constant P=RTV VP2= constant ⇒ V×R2T2V2= constant R2T2V= constant ⇒ T2V= Constant T2V= T'22V or T'2=2T2 T' = 2T

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

During an experiment an ideal gas is found to obey an additional law VP² = constant. The gas is initially at temperature T and volume V, when it expands to volume 2 V, the resulting temperature is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{T}{2}$

$2 T$

$\sqrt{2} T$

$\frac{T}{\sqrt{2}}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P V = R T = c o n s \tan t P = \frac{R T}{V} V P^{2} = c o n s \tan t \Rightarrow \frac{V \times R^{2} T^{2}}{V^{2}} = c o n s \tan t \frac{R^{2} T^{2}}{V} = c o n s \tan t \Rightarrow \frac{T^{2}}{V} = C o n s \tan t \frac{T^{2}}{V} = \frac{T'^{2}}{2 V} o r T'^{2} = 2 T^{2} T' = \sqrt{2} T$