Answer to Question #84275 in Chemistry for Sayem

Question

In SATP to increase 1 mole gas volume to 40L, how much temperature should increase if pressure is half decreased---
1.389.43K
2.91.43°C
3.298K
4.116.43°C

Answer 1

Find volume of 1 mole of a gas at SATP:

P= 101325 Pa

n= 1 mole

R= 8.314 m³Pa/(K*mole)

T = 298.15

PV=nRT

V= nRT/V = 1*8.314*298.15/101325 = 24.5 dm³

Use Combined Gas Law to determine final temparature:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁/2

P₁*24.5 /298.15 = (P₁/2) * 40 / T₂

24.5/298.15 = 20/T₂

T₂ = 243.39 K

So, we can see that there is no increase in temperature as it is said in the task. There is decrese in temperature: T₁ = 298.15 K and T₂ = 243.39 K. The answer is: none of the above. Or there should be a mistake in the very task.