if 429 grams of gas occupies 25.6L at 49 degrees C at a pressure of 759 mmHg what volume will 4.29g of gas will occupy at 53 degrees C (assume pressure is unchanged)

The long way.

PV = nRT
Solve for n, then n = g/molar mass. Solve for molar mass.
Redo PV = nRT, solve for V under the new conditions.

Easier way.
42.9 is 0.1 of the initial so at the same T and P, V = 0.1 x 25.6 or 2.56 L.
Then correct for the temperature difference.
2.56 x 326/322 = ??

lmk

To solve this problem, we will use the Ideal Gas Law equation, which is stated as:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal Gas Constant (0.0821 L·atm/mol·K for this problem)
T = Temperature in Kelvin

First, we need to convert the given temperature from Celsius to Kelvin.
To convert Celsius to Kelvin, we add 273.15 to the temperature in Celsius.
So, 49 degrees Celsius + 273.15 = 322.15 Kelvin.

To find the number of moles of gas, we need to use the given mass and the molar mass of the gas.
The molar mass can be determined by looking up the atomic masses of the elements in the gas and adding them up. Since we don't have that information, we'll assume the gas is an ideal gas for simplicity.

Given mass of gas = 429 grams
Molar mass of gas = unknown (for simplicity, we'll assume it's 1g/mol)

Number of moles of gas = mass of gas / molar mass of gas
Number of moles of gas = 429g / 1g/mol = 429 moles

Now, we need to find the volume of gas when the conditions change.

Temperature = 53 degrees Celsius + 273.15 = 326.15 Kelvin (converted to Kelvin)

Using the Ideal Gas Law equation, we can rearrange it to solve for V2:

V1 / T1 = V2 / T2

Where:
V1 = Initial volume (25.6L)
T1 = Initial temperature in Kelvin (322.15K)
V2 = Final volume (what we want to find)
T2 = Final temperature in Kelvin (326.15K)

Plugging in the values:

25.6L / 322.15K = V2 / 326.15K

Now, we can solve for V2:

V2 = (25.6L * 326.15K) / 322.15K
V2 = 26.04L

Therefore, 4.29 grams of gas will occupy approximately 26.04 liters at 53 degrees Celsius (assuming the pressure remains unchanged).