A gas occupies 475 mL at a pressure of 115.8 kPa and a temperature 22 degrees celsius. The pressure on the gas is increased to 143 kPa and the volume is decreased to 288 mL. What is the new temperature in degrees celsius? (RECORD ANSWER TO NEAREST WHOLE NUMBER

-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

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