Answer:
-52°C is the new temperature in degrees celsius
Explanation:
P . V = n . R . T
That's the formula for the Ideal Gases Law, that can be applied here. As n (number of moles) and R, are the same number we can cancel them, for both situations so, finally we have:
P₁ . V₁ / T₁ = P₂ . V₂ / T₂
We do some conversions, before:
475 mL . 1L / 1000 mL = 0.475L
115.8 kPa . 1 atm / 101.3 kPa = 1.14 atm
22°C + 273 = 295K
143 kPa . 1 atm / 101.3 kPa = 1.41 atm
288 mL . 1L/1000 mL = 0.288L
We replace: (1.14 atm . 0.475L) / 295K = (1.41 atm . 0.288L) / T₂
1.83×10⁻³ atm.L/K . T₂ = 1.41 atm . 0.288L
T₂ = 1.41 atm . 0.288L / 1.83×10⁻³ K/atm.L
T₂ = 221.2 K
Let's convert the °C value = 221.2 K - 273 = -51.8°C ≅ 52°C
Answer:
The new temperature is -52°C
Explanation:
Step 1: Data given
Volume of the gas = 475 mL = 0.475 L
Pressure of a gas = 115.8 kPa
Temperature = 22 °C = 295 K
The pressure of the gas is increased to 143 kPa
The volume is decreased to 288 mL
Step 2: Calculate the new temperature
(P1*V1)/T1 = (P2*V2)/T2
⇒with P1 = the initial pressure = 115.8 kPa
⇒with V1 = the initial volume = 0.475 L
⇒with T1 = the initial temperature = 295 K
⇒with P2 the increased pressure = 143 kPa
⇒with V2 = the decreased volume = 0.288 L
⇒with T2 = the new temperature = TO BE DETERMINED
(115.8 * 0.475) / 295 = (143 * 0.288)/T2
T2 = (143*0.288) / ((115.8 * 0.475) / 295)
T2 = 221 K
Step 3: Convert temperature from Kelvin to celsius
°C = Kelvin - 273
°C = 221 - 273 = -52 °C
The new temperature is -52°C
