Question

It is found that on heating a gas its volume increases by 50% and its pressure decreases to
60% of its original value. If the original temperature was -15°C, find the temperature to which it was
heated.
​

Answer 1

Answer:

ideal gas equation-

PV=nRT(where P is pressure,V is volume,n is the no. of moles present,R is the gaseous constant and T is the temperature)

• P,V and T are changing...and moles of gas is constant here...as its not escaping
• so we can establish the relation as ...
• $\frac{pv}{t} = nr = constant$
• $\frac{p_{1}v_{1} }{t1} = \frac{p2v2}{t2}$
• P1=P...V1=V...T1=-15°C=258K(☆NOTE =put all quantities in standard units)

here according to ques..

☆pressure decreases to 60%

---- P2 =6P/10

☆volume increases by 50%

-----V2=V+V/2

=3V/2

...lets solve...

put the given values in the equation...

• $\frac{p \times v}{258} = \frac{6p \times 3v}{10 \times .2 \times .t}$
• $t = \frac{258 \times 9}{10}$
• T2=232.2K
• T2=-19.8°C

hope it helps

Good luck ♧♧