Question

68. 2.8g of N, gas at 300 K and 20 atm was
allowed to expand isothermally against a
constant external pressure of 1 atm. Calculate
W for the gas.
1) +236.95 J
2) + 136.95 J
3) - 236.95 J
4) + 136.95 J​

Answer 1

Answer:

Consult it on wikipidia

Answer 2

Value of W for the given gas is -236.95 J.

Explanation:

The given data is as follows.

         mass = 2.8 g,      T = 300 K,  

      $P_{1}$ = 20 atm,   and    $P_{1}$ = 1 atm

Hence, first we will calculate the number of moles of nitrogen gas as follows.

          No. of moles = $\frac{mass}{\text{molar mass}}$

                                = $\frac{2.8 g}{28 g/mol}$

                                = 0.1 mol

Therefore, calculate the work done as follows.

          W = $-P_{ext} nRT log (\frac{1}{P_{2}} - \frac{1}{P_{1}})$

              = $-1 atm \times 0.1 mol \times 8.314 J/mol \times 300 K log (\frac{1}{1} - \frac{1}{20})$

              = -236.95 J

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