Question

hii guys.


calculate the work done if0.75mole of an ideal gas expands at 300k from volume of 15ltrs to 25ltrs. given R=8.314


solve this question please.....​

Answer 1

Explanation:

Answer: The value of W, Q and dU for the given process are -955.75 J, -955.75 J and 0 J respectively.

Explanation:

To calculate the work done for isothermal, reversible expansion process, we use the equation:

where,

W = work done

n = number of moles = 0.75 mole

R = Gas constant = 8.314 J/mol.K

T = Temperature of the gas =

= initial volume = 15 L

= final volume = 25 L

Putting values in above equation, we get:

The equation for first law of thermodynamics follows:

where,

Q = total amount of heat required = ?

dU = Change in internal energy = 0 (for isothermal process)

W = work done

So, Q = W = -955.75 J

Hence, the value of W, Q and dU for the given process are -955.75 J, -955.75 J and 0 J respectively.

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Answer 2

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