Question

A set sample of gas has an initial volume of 445 mL and a pressure of 2.5 atm. The gas is changed to STP (Standard Temperature and Pressure), and the new volume is 289 mL. What was the original temperature in °C?

Answer 1

You can use the ideal gas law, which is expressed as:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (approximately 0.0821 L atm / mol K)
T = temperature (in Kelvin)

First, let's find the initial number of moles (n) using the initial conditions:

Initial volume (V1) = 445 mL = 0.445 L
Initial pressure (P1) = 2.5 atm

Rearrange the ideal gas law to solve for n:

n = (P1 * V1) / (R * T1)

Now, we need to convert the final conditions to STP (Standard Temperature and Pressure), where STP is defined as 0°C (273.15 K) and 1 atm. So:

Final volume (V2) = 289 mL = 0.289 L
Final pressure (P2) = 1 atm
T2 = 273.15 K (STP temperature)

Now, let's calculate n again using the final conditions:

n = (P2 * V2) / (R * T2)

Now, set the two expressions for n equal to each other since the number of moles doesn't change:

(P1 * V1) / (R * T1) = (P2 * V2) / (R * T2)

Now, plug in the values:

(2.5 atm * 0.445 L) / (0.0821 L atm / mol K * T1) = (1 atm * 0.289 L) / (0.0821 L atm / mol K * 273.15 K)

Now, solve for T1:

T1 = (2.5 atm * 0.445 L * 273.15 K) / (1 atm * 0.289 L)

T1 ≈ 1021.94 K

To convert this temperature to degrees Celsius:

T1 in °C ≈ 1021.94 K - 273.15 ≈ 748.79°C

So, the original temperature in degrees Celsius was approximately 748.79°C.