A 500.0 mL gas sample at STP is compressed to a volume of 300.0 mL and the temperature is increased to 35.0°C. What is the new pressure of the gas in pascals?
To find the new pressure of the gas, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in pascals)
V = volume (in cubic meters)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the given volume from milliliters to cubic meters:
300.0 mL ÷ 1000 mL = 0.3 L ÷ 1000 L = 0.0003 m^3
Next, we need to convert the temperature from Celsius to Kelvin. The Kelvin scale is used in the ideal gas law equation. The conversion formula is:
T(K) = T(°C) + 273.15
T(K) = 35.0°C + 273.15 = 308.15 K
Now, we can rearrange the ideal gas law equation to solve for P:
P = (nRT) / V
At STP (standard temperature and pressure), one mole of gas occupies 22.4 liters. Therefore, the number of moles (n) is equal to the initial volume (500.0 mL) divided by 1000 mL/L (to convert mL to liters) divided by 22.4 L/mol:
n = (500.0 mL ÷ 1000 mL/L) ÷ 22.4 L/mol = 0.0223214286 mol
Let's substitute the values into the equation:
P = (0.0223214286 mol × 8.314 J/(mol·K) × 308.15 K) / 0.0003 m^3
P ≈ 564,457.24 Pa
Therefore, the new pressure of the gas is approximately 564,457.24 pascals.
what are P and T (in °K) at STP?
PV=kT, so PV/T = k is constant
so just find the new P such that PV/T remains unchanged from at STP