need help with the picture portion (part 4 of 7 what is the final pressure Pf of

Question

need help with the picture portion (part 4 of 7 what is the final pressure Pf of the gas). But the info below goes with the picture

A 0.409-L canister contains a diatomic ideal gas at temperature 305 K and pressure 3.50 ? 105 Pa. If the gas is suddenly released and expands to a volume of 0.465 L, determine the following.

(a) Determine the final pressure of the gas.

Part 4 of 7 Identify the thermodynamic process. (cont.) To solve part (a), because no thermal energy transfers into or out of the gas, the process is adiabatic and Pvr = constant, where = is the adiabatic index of the gas. For an ideal diatomic gas, cp = 2R and Cy- R so that C. 5 2 7 8 7 SR5 The gas pressures and volumes in the initial and final states are related by the equation P· = PN/ What is the final pressure Pr of the gas? Pa

Explanation / Answer

we know, for diatomic gas, gamma = 1.4

V1 = 0.409 L

P1 = 3.5*10^5 pa

V2 = 0.465 L

P2 = ?

In adiabatic process, P*V^gamma = constant.

so Apply, P1*V1^gamma = P2*V2^gamma

P2 = P1*(V1/V2)^gamma

= 3.5*10^5*(0.409/0.465)^1.44

= 2.91*10^5 pa <<<<<<<<<<<------------Answer

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