Question

To what temperature must a neon gas sample be heated to double its pressure if the initial volume of gas at 75°c is decreased by 15.0%

Answer 1

We know, PV =nRT
And n,R are constants.
Hence we can write
P1*V1/T1=P2*V2/T2
Here,
T1=273+75= 348 K ,P2=2P1,
V2=V1-15V1/100 = 0.85V1, T2= ?

Now substitute the values in the derived equation;
P1*V1/348 = 2P1*0.85V1/T2
1/348 = (2*0.85)/T2
T2 = 348*1. 7 = 591.6 K

If in °C then,591.6 - 273 = 318.6°C

Hope it is helpful....

Answer 2

By ideal gas equation,

P1.V1/T1 = P2.V2/T2

T1 = 273 + 75 = 348K
P2 = 2P1
Initial volume = V1
V2 = V1 - 15V1/100 = 0.85V1
T2 = ?

P1.V1/348 = 2P1 × 0.85V1/T2

1/348 = 2 × 0.85/T2
T2 = 348 × 1/7 = 591.6K
In degrees, 591.6 - 273 = 318.6°C