Question

A sample of an ideal gas has a volume of 2.31 L

at 279 K
and 1.01 atm.
Calculate the pressure when the volume is 1.09 L
and the temperature is 308 K.

Answer 1

To find the pressure when the volume and temperature change, we can use the ideal gas law equation.

Step-by-step explanation:

To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to convert the given temperatures to Kelvin by adding 273.15. The initial temperature is 279 K, and the final temperature is 308 K.

Next, we can set up a ratio using the initial and final volumes:

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

Plugging in the given values:

(1.01 atm * 2.31 L) / 279 K = (P2 * 1.09 L) / 308 K

Solving for P2:

P2 = ((1.01 atm * 2.31 L) / 279 K) * (308 K / 1.09 L)

P2 is approximately equal to 2.38 atm.

Answer 2

We can use the combined gas law to determine the pressure of the gas at the final state. The combined gas law relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.

We are given the initial pressure (P1 = 1.01 atm), volume (V1 = 2.31 L), and temperature (T1 = 279 K) of the gas, and the final volume (V2 = 1.09 L), and temperature (T2 = 308 K) of the gas. We can solve for P2, the final pressure of the gas:

(P1 x V1) / T1 = (P2 x V2) / T2

P2 = (P1 x V1 x T2) / (V2 x T1)

P2 = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)

P2 = 2.41 atm (rounded to three significant figures)

Therefore, the pressure of the gas when the volume is 1.09 L and the temperature is 308 K is approximately 2.41 atm.