Question

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Answer 1

P¹V¹/T¹=P²V²/T² here

P²= 1.25- atm

V²=?

T²=345K

P¹=0.75-atm

V¹=120

T¹=295K

So V²= (0.75 x120/295)x(345/1.25)

Volume of the gas V²= 84.20- ml

Answer 2

Solution given:

at first

volume
`V_(1)=120.0ml`

pressure
`P_(1)=0.75atm`

temperature
`T_(1)=22+273=295°K`

after increasing temperature and pressure

volume
`V_(2)=xml`

pressure
`P_(2)=1.25atm`

temperature
`T_(2)=70+273=343°K`

now

By using Combined gas equation

`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`

substituting value

`(0.75*120)/(295)=(1.25*x)/(343)`

0.305=5x/1372

Doing criss cross multiplication

5x=0.305*1372

x=418.46/5

x=83.69ml

x=84ml

the volume will be 84ml