a 0.25 mol sample af neon gas at 18 C and 0.700 atm is heated to 40 C. the new gas pressure is 1.11 atm. what is the change in volumn of the neon gas?
(V1/T1) = (V2/T2)
To find the change in volume of the neon gas, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(40°C) = 40 + 273.15 = 313.15 K
T(18°C) = 18 + 273.15 = 291.15 K
Next, we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
We are given the following information:
Initial pressure, P1 = 0.700 atm
Initial temperature, T1 = 291.15 K
Final pressure, P2 = 1.11 atm
Final temperature, T2 = 313.15 K
Number of moles, n = 0.25 mol
Now let's calculate the initial volume, V1:
V1 = (nRT1) / P1
Substituting the values:
V1 = (0.25 mol * 0.0821 L·atm/mol·K * 291.15 K) / 0.700 atm
V1 ≈ 27.60 L
Now we can calculate the final volume, V2, using the same equation:
V2 = (nRT2) / P2
Substituting the values:
V2 = (0.25 mol * 0.0821 L·atm/mol·K * 313.15 K) / 1.11 atm
V2 ≈ 6.99 L
Finally, we can find the change in volume by subtracting the initial volume from the final volume:
Change in volume = V2 - V1
Change in volume ≈ 6.99 L - 27.60 L
Change in volume ≈ -20.61 L
Therefore, the change in volume of the neon gas is approximately -20.61 liters. Note that the negative sign indicates a decrease in volume.