pV = nRT
p = nRT/V
p= 1 x 0.08205 x 1000/ 2
p = 41.025 Pa
Edit: The unit should be atm instead of Pa, as pointed out by a nice human being.
Answer:
No, there is no evidence that the manufacturer has a problem with underfilled or overfilled bottles, due that according our results we cannot reject the null hypothesis.
Explanation:
according to this exercise we have the following:
σ^2 =< 0.01 (null hypothesis)
σ^2 > 0.01 (alternative hypothesis)
To solve we can use the chi-square statistical test. To reject or not the hypothesis, we have that the rejection region X^2 > 30.14
Thus:
X^2 = ((n-1) * s^2)/σ^2 = ((20-1)*0.0153)/0.01 = 29.1
Since 29.1 < 30.14, we cannot reject the null hypothesis.
First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps
2.24 liters is the volume of the gas if pressure is increased to 1000 Torr.
Explanation:
Data given:
Initial volume of the gas V1 = 2.6 liters
Initial pressure of the gas P1 = 860 Torr 1.13 atm
final pressure on the gas P2 = 1000 Torr 1.315 atm
final volume of the gas after pressure change V2 =?
From the data given above, the law used is :
Boyles Law equation:
P1V1 = P2V2
V2 = P1V1/P2
= 1.13 X 2.6/ 1.31
= 2.24 Liters
If the pressure is increased to 1000 Torr or 1.315 atm the volume changes to 2.24 liters. Initially the volume was 2.6 litres and the pressure was 860 torr.
Answer:
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
Explanation:
Data Given:
Moles = n = 3.2 mol
Temperature = T = 312 K
Pressure = P = ?
Volume = V = 87 m³ = 87000 L
Formula Used:
Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for P,
P = n R T / V
Putting Values,
P = (3.2 mol × 0.082057 atm.L.mol⁻¹.K⁻¹ × 312 K) ÷ 87000 L
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
