Question

A cylinder contains 0.100mol of an ideal monatomic gas. Initially the gas is at a pressure of 1.00×105Pa and occupies a volume of 2.50×10−3m3. A) Find the initial temperature of the gas in kelvins. B)If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) of the gas if the expansion is isothermal. C)Find the final pressure of the gas in this process. D)If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) of the gas if the expansion is isobaric. E)Find the final pressure of the gas in this process. F)If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) of the gas if the expansion is adiabatic. G)Find the final pressure of the gas in this process.

Answer 1

A) The initial temperature of the gas is 299.48 K. B) If the expansion process is isothermal, the final temperature will be 299.48 K as well. C) The final pressure of the gas in an isothermal process is 5.00 x 10^4 Pa.

Step-by-step explanation:

A) To find the initial temperature of the gas in kelvins, you 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.

Plugging in the given values:

1.00 x 10^5 Pa * 2.50 x 10^-3 m^3 = 0.100 mol * R * T

Arranging the equation and solving for T:

T = (P * V) / (n * R) = (1.00 x 10^5 Pa * 2.50 x 10^-3 m^3) / (0.100 mol * 8.314 J/(mol·K)) = 299.48 K

Therefore, the initial temperature of the gas is 299.48 kelvins.

B) If the gas is allowed to expand to twice the initial volume and the process is isothermal, the temperature will remain constant. Therefore, the final temperature of the gas will be the same as the initial temperature, which is 299.48 kelvins.

C) In an isothermal process, the final pressure is given by:

P1 * V1 = P2 * V2

Plugging in the given values:

(1.00 x 10^5 Pa) * (2.50 x 10^-3 m^3) = P2 * (2.50 x 10^-3 m^3 * 2)

Solving for P2:

P2 = (1.00 x 10^5 Pa * 2.50 x 10^-3 m^3) / (2.50 x 10^-3 m^3 * 2) = 5.00 x 10^4 Pa

Therefore, the final pressure of the gas in this process is 5.00 x 10^4 Pa.

D) If the gas is allowed to expand to twice the initial volume and the process is isobaric, the initial pressure remains constant. Therefore, the final temperature can be found using the equation:

V1 / T1 = V2 / T2

Plugging in the given values:

(2.50 x 10^-3 m^3) / (299.48 K) = (2.50 x 10^-3 m^3 * 2) / T2

Solving for T2:

T2 = (2.50 x 10^-3 m^3 * 2) / ((2.50 x 10^-3 m^3) / (299.48 K)) = 2 * 299.48 K = 598.96 K

Therefore, the final temperature of the gas in this process is 598.96 kelvins.

E) In an isobaric process, the final pressure remains constant. Therefore, the final pressure of the gas in this process is the same as the initial pressure, which is 1.00 x 10^5 Pa.

F) If the gas is allowed to expand to twice the initial volume and the process is adiabatic, there is no heat transfer. Therefore, the final temperature can be found using the equation:

T2 = T1 * (V1 / V2) ^ ((y - 1) / y)

Plugging in the given values and the value of y = 5/3 for monatomic gas:

T2 = 299.48 K * (2.50 x 10^-3 m^3 / (2.50 x 10^-3 m^3 * 2))^((5/3 - 1) / (5/3)) = 87.45 K

Therefore, the final temperature of the gas in this process is 87.45 kelvins.

G) In an adiabatic process, the final pressure can be found using the equation:

P2 = P1 * (V1 / V2) ^ y

Plugging in the given values and the value of y = 5/3 for monatomic gas:

P2 = 1.00 x 10^5 Pa * (2.50 x 10^-3 m^3 / (2.50 x 10^-3 m^3 * 2))^(5/3) = 4.00 x 10^4 Pa

Therefore, the final pressure of the gas in this process is 4.00 x 10^4 Pa.

Answer 2

Calculating the initial temperature, final temperature, and pressure of an ideal gas involves using the ideal gas law and additional relationships for specific processes such as isothermal, isobaric, and adiabatic expansions.

Step-by-step explanation:

Finding the Temperature and Pressure of an Ideal Gas under Different Conditions

To solve this physics problem regarding the states of an ideal gas under different conditions, we will employ the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvins. For adiabatic processes, we also use the relationship PVγ is constant, where γ (gamma) is the specific heat capacity ratio.

Part A - Initial Temperature Calculation

We start by finding the initial temperature (Ti) using the ideal gas law:

PV = nRT

Ti = (P * V) / (n * R)

Using the given values, we can now calculate Ti.

Part B & C - Isothermal Expansion

Since the expansion is isothermal, the temperature remains constant, and the final pressure can be found using the equation:

P1V1 = P2V2

We'll solve for P2 with V2 being twice V1.

Part D - Isobaric Expansion Temperature

The final temperature under isobaric (constant pressure) conditions can be calculated as follows:

V1/T1 = V2/T2

Calculate T2 using the above relation.

Part E - Isobaric Expansion Pressure

During isobaric expansion, the pressure remains constant, so Pfinal = Pinitial.

Part F & G - Adiabatic Expansion

For adiabatic expansion, we use:

P1V1γ = P2V2γ for pressure, and T1V1γ-1 = T2V2γ-1 for temperature.

Using these relations and the given initial conditions, calculate the final pressure and temperature.