Correct the following samples of gases to standard temperature. (Assume pressure is constant)(STP stands for standard temp and pressure which is 273K and 101.3 kPa, you do not have to think about the pressure for these four but on the last three you will have to so T2 is 273K and on the last three problems for this assignment P2 will be 101.3 kPa)

Question

Correct the following samples of gases to standard temperature. (Assume pressure is constant)(STP stands for standard temp and pressure which is 273K and 101.3 kPa, you do not have to think about the pressure for these four but on the last three you will have to so T2 is 273K and on the last three problems for this assignment P2 will be 101.3 kPa)

1. 1622 cm3 at 15.0 oC

2. 12.33 L at 280 K

3. 155.6 dm3 at 110 K 4. 1250 cm3 at -45.0 oC

Answer 1

I don't know which are the firs three and the last four.

Answer 2

To correct the given samples of gases to standard temperature, you need to use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (8.31 J/mol·K), and T is the temperature in Kelvin.

To solve these problems, we will use the equation and rearrange it to solve for the corrected volume at standard temperature (V₂). Since pressure is assumed to be constant, we can rewrite the equation as:

(V₁/T₁) = (V₂/T₂)

1. Sample: 1622 cm³ at 15.0 °C
To convert temperature to Kelvin, add 273.
T₁ = 273 + 15 = 288 K

Now we can rearrange the equation and solve for V₂:
(1622 cm³ / 288 K) = (V₂ / 273 K)

V₂ = (1622 cm³ * 273 K) / 288 K
V₂ ≈ 1535.24 cm³

Therefore, the corrected volume at standard temperature is approximately 1535.24 cm³.

2. Sample: 12.33 L at 280 K
T₁ = 280 K

(12.33 L / 280 K) = (V₂ / 273 K)

V₂ = (12.33 L * 273 K) / 280 K
V₂ ≈ 12.04 L

The corrected volume at standard temperature is approximately 12.04 L.

3. Sample: 155.6 dm³ at 110 K
To convert volume from dm³ to cm³, multiply by 1000.
V₁ = 155.6 dm³ * 1000 cm³/dm³ = 155600 cm³
T₁ = 110 K

(155600 cm³ / 110 K) = (V₂ / 273 K)

V₂ = (155600 cm³ * 273 K) / 110 K
V₂ ≈ 386889.09 cm³ (or 386.89 L, if you prefer)

The corrected volume at standard temperature is approximately 386.89 L.

4. Sample: 1250 cm³ at -45.0 °C
T₁ = -45.0 °C + 273 = 228 K

(1250 cm³ / 228 K) = (V₂ / 273 K)

V₂ = (1250 cm³ * 273 K) / 228 K
V₂ ≈ 1494.74 cm³

The corrected volume at standard temperature is approximately 1494.74 cm³.

Remember to always convert temperature to Kelvin and use the appropriate units in calculations.