Answer:
The volume decreases 5.5%
Explanation:
First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.
Now, with this data, let's see what we can do.
If this is an ideal gas, the equation to use is:
PV = nRT
Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:
n₁ = n₂ = n
The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:
P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁
Doing the same with the pressure and volume 2 we have:
n = P₂V₂ / RT₂
Equalling both expressions and solving for V₂:
P₁V₁ / RT₁ = P₂V₂ / RT₂
V₂ = P₁T₂V₁ / P₂T₁
Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:
V₂ = P₁T₂V₁ / 0.95P₁T₁
The values of temperature in K:
T1 = 21+273 = 294 K
T2 = -8 + 273 = 265 K
Finally, let's calculate the volume:
V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out
V₂ = 264V₁ / 294*0.95
V₁ = 0.945V₂
With this, we can day that Volume 2 decreases.
Now the percentage change would be using the following expression:
%V = (V₁ - V₂ / V₁) * 100
Replacing the data we have:
%V = V1 - 0.945V₁ / V₁
%V = 0.055V₁ / V₁ * 100
%V = 5.5%
= 0.668 mol Ni