Question

A 3.35 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1570 m 3 to 0.770 m 3 . Initially the pressure was 1.00 atm. Determine the initial and final temperatures. Enter your answers numerically separated by a comma.

Answer 1

To determine the initial and final temperatures of the gas, we can use the adiabatic expansion equation:

P₁V₁^γ = P₂V₂^γ

Where P₁ and P₂ are the initial and final pressures respectively, V₁ and V₂ are the initial and final volumes, and γ is the heat capacity ratio of the gas. For a diatomic gas, γ is equal to 1.4.

First, let's find the final pressure using the ideal gas law equation:

P₁V₁/T₁ = P₂V₂/T₂

Since the expansion is adiabatic, there is no heat exchange, which means the change in temperature (T₂ - T₁) is zero. Rearranging the equation, we get:

P₂ = (P₁V₁/V₂)

Now, let's substitute this value for P₂ into the adiabatic expansion equation:

(P₁V₁^γ) = (P₁V₁/V₂)^γ

Simplifying, we get:

V₁^γ = V₁^γ / V₂^γ

Taking the ratio of the volumes:

(V₁/V₂) = (V₁^γ / V₂^γ)^(1/γ)

Since γ = 1.4 for a diatomic gas, we have:

(V₁/V₂) = (V₁^1.4 / V₂^1.4)^(1/1.4)

Now, let's substitute the given values:

(V₁/V₂) = (0.1570^1.4 / 0.770^1.4)^(1/1.4)

Using a calculator, we find:

(V₁/V₂) ≈ 0.4244

Next, let's find the final pressure:

P₂ = (P₁V₁/V₂)

P₂ = (1.00 atm)(0.4244)

P₂ ≈ 0.4244 atm

Now, we can find the initial and final temperatures using the ideal gas law equation:

P₁V₁/T₁ = P₂V₂/T₂

Substituting the known values:

(1.00 atm)(0.1570 m^3)/T₁ = (0.4244 atm)(0.770 m^3)/T₂

Simplifying, we get:

T₂ = (T₁)(0.1570 m^3)(0.4244 atm)/(0.770 m^3)(1.00 atm)

T₂ ≈ (T₁)(0.0838)

Since the change in temperature is zero, we have:

T₁ ≈ T₂

Therefore, the initial and final temperatures are approximately the same. We can represent this as:

T₁, T₂ ≈ T

So, the answer is: T, T (where T represents the temperature).