Answer:
2. 181.25 K.
3. 0.04 atm.
Explanation:
2. Determination of the temperature.
Number of mole (n) = 2.1 moles
Pressure (P) = 1.25 atm
Volume (V) = 25 L
Gas constant (R) = 0.0821 atm.L/Kmol
Temperature (T) =?
The temperature can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
1.25 × 25 = 2.1 × 0.0821 × T
31.25 = 0.17241 × T
Divide both side by 0.17241
T = 31.25 / 0.17241
T = 181.25 K
Thus, the temperature is 181.25 K.
3. Determination of the pressure.
Number of mole (n) = 10 moles
Volume (V) = 5000 L
Temperature (T) = –10 °C = –10 °C + 273 = 263 K
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =?
The pressure can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
P × 5000 = 10 × 0.0821 × 263
P × 5000 = 215.923
Divide both side by 5000
P = 215.923 / 5000
P = 0.04 atm
Thus, the pressure is 0.04 atm
Answer:
d = 0.992 g/L
Explanation:
Data Given:
Pressure of nitric oxide (NO) = 0.866 atm
Temperature of a gas = 46.2° C
Convert the temperature to kelvin = 46.2° C + 273
temperature in kelvin = 319.2 K
density of nitric oxide (NO) = ?
Solution:
Density of a gas can be calculated by
d = PM /RT
Where
d = density
P = Pressure
M = molar mass of gas
R = ideal gas constant = 0.0821 L atm mol⁻¹ K⁻¹
T = temperature
So,
Molar mass of NO = 30 g/mol
Put values in the formula:
d = PM /RT
d = 0.866 atm × 30 g/mol / 0.0821 L atm mol⁻¹ K⁻¹ × 319.2 K
d = 25.98 atm. g/mol / 26.2 L atm mol⁻¹
d = 0.992 g/L
Answer:
354.67K
Explanation:
Applying
P₁V₁/T₁ = P₂V₂/T₂................. Equation 1
Where Where P₁ = initial pressure, T₁ = Initial temperature, V₁ = Initial Volume, P₂ = Final pressure, V₂ = Final Volume, T₂ = Final Temperature.
From the question, we are ask to look for the final temperature,
Therefore we make T₂ the subject of the equation
T₂ = P₂V₂T₁/P₁V₁............. Equation 2
Given: P₁ = 600 kPa, V₁ = 500 mL, T₁ = 77 °C = (273+77) = 350 K, P₂ = 760 kPa, V₂ = 400.0 mL
Substitute these values into equation 2
T₂ = (760×400×350)/(600×500)
T₂ = 354.67 K
