Suppose that subsequent to this heating, 1.30 104 J of thermal energy is removed

Question

Suppose that subsequent to this heating, 1.30 104 J of thermal energy is removed from the gas isothermally. Find the final volume in terms of the initial volume of the example, V0. (Hint: Follow the same steps as in the example, but in reverse. Use the values calculated in the solution portion. Also note that the initial volume in this exercise is 1.25 V0.)

Follow the example, but do the steps in reverse order. Make sure you know how to solve for x in ln(x) =

This is the example-i did this but i got .93. I am probably making a calculator mistake so please give me the answer so that i can try to figure out what im doing wrong. thanks!

Substitute into the isothermal work equation, finding the work done during the isothermal expansion. Note that T = 27.0°C = 3.00 102 K. Wenv = nRT ln (Vf/Vi)
= (5.00 mol)(8.31 J/K · mol)(3.00 102 K) ln ((1.25 V0)/ V0)
Wenv = 2.78 103 J
Q = Wenv = 2.78 103 J .The negative of this amount is the work done on the gas. W = -Wenv = -2.78 103 J ..

Explanation / Answer

thermal enenrgy removed Q = 1.3 * 10 ^ 4 J when heat is removed then it taken as negative So, Q = -1.3 * 10 ^ 4 J for isothermal process W = Q = -1.3 * 10 ^ 4 J we know W = nRT ln ( Vf / Vi) -1.3 * 10 ^ 4 = nRT ln ( Vf / Vi ) 1.3 * 10 ^ 4 = nRT ln ( Vi / Vf ) Since ln( Vi / Vf ) = -ln ( Vf / Vi ) from this you find Vi / Vf value

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