Answer:
there will be less and less space
Explanation:
Answer:
The final temperature is 566°K
Explanation:
Given
V1 = 0.444L
T1 = 0.00°C
P1 = 79.00 kPa
V2 = 1880mL
P2 = 38.70 kPa
Required
Determine the final temperature of the gas
To answer this, we make use of idea gas law equation.
This is:
(P1V1)/T1 = (P2V2)/T2
Convert V1 to mL
V1 = 0.444L
V1 = 0.444 * 1000mL
V1 = 444mL
Convert T1 to Kelvin
T1 = 0.00°C
T1 = 0.00 + 273K
T1 = 273K
Substitute these values in the given equation.
(P1V1)/T1 = (P2V2)/T2
(79 * 444)/273 = (38.70 * 1880)/T2
35076/273 = 72756/T2
Cross Multiply
35076 * T2 = 273 * 72756
35076T2 = 19862388
Make T2 the subject
T2 = 19862388 ÷ 35076
T2 = 566K (approximated)
Hence, the final temperature is 566°K
Answer:
∴ The absolute pressure of the air in the balloon in kPa = 102.69 kPa.
Explanation:
- We can solve this problem using the general gas law:
<em>PV = nRT</em>, where,
P is the pressure of the gas <em>(atm)</em>,
V is the volume of the gas in L <em>(V of air = 6.23 L)</em>,
n is the no. of moles of gas <em>(n of air = 0.25 mole)</em>,
R is the general gas constant <em>(R = 0.082 L.atm/mol.K)</em>,
T is the temperature of gas in K <em>(T = 35 °C + 273 = 308 K</em>).
∴ P = nRT / V = (0.25 mole)(0.082 L.atm/mol.K)(308 K) / (6.23 L) = 1.0135 atm.
- <em>Now, we should convert the pressure from (atm) to (kPa).</em>
1.0 atm → 101.325 kPa,
1.0135 atm → ??? kPa.
∴ The absolute pressure of the air in the balloon in kPa = (101.325 kPa)(1.0135 atm) / (1.0 atm) = 102.69 kPa.
Answer:
kinetic energy
Explanation:
anything moving contains kinetic energy and anything standing still has potential.
Explanation:
IR (infrared) spectroscopy theory is based on the concept that the molecules tend to absorb some specific amount of the frequencies of light which are characteristic of corresponding structure of molecules. These energies are reliant on shape of molecular surfaces, associated vibration coupling, and mass corresponding to atoms.
It is used to detect the type of the functional group present in the organic compound.