5 moles of Hydrogen y -7-5 initially at S-T-P- are compressed adiabatically so that its temperature becomes 400- C- The increase in the internal energy of the gas in kilo - joules is- R -8-30 J mol -1 K -1A- 65-55B- 80-55C- 21-55D- 41-5

The correct option is D 41.5Given T_1 = 0^oC = 273 K, T_2 = 400^oC = 673 K Work done W = nR/(γ 1)(T_2 T_1) = 5 × 8.3 × 400/( 7/5 1 ) = 41500 J = 41.5 kJ ...

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Standard XII

Physics

1st Law of Thermodynamics

5 moles of Hy...

Question

5 moles of Hydrogen $(γ = \frac{7}{5})$ initially at S.T.P. are compressed adiabatically so that its temperature becomes $400^{o} C$ . The increase in the internal energy of the gas in kilo - joules is: $(R = 8.30 J m o l^{- 1} K^{- 1})$

21.55

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41.5

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65.55

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80.55

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Solution

The correct option is B 41.5
Given $T_{1} = 0^{o} C = 273 K, T_{2} = 400^{o} C = 673 K$
Work done $W = \frac{n R}{(γ - 1)} (T_{2} - T_{1}) = \frac{5 \times 8.3 \times 400}{(\frac{7}{5} - 1)} = 41500 J = 41.5 k J$
By convention, the work done on the gas is taken to be negative, i.e. W = - 41.5 kJ. From the first law of thermodynamics dQ = dU + dW. For an adiabatic process, dQ = 0. Hence dU = - dW = - ( - 41.5 ) = 41.5 kJ. The positive sign of dU implies that the internal energy increases.

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5 moles of Hydrogen (γ=75) initially at S.T.P. are compressed adiabatically so that its temperature becomes 400oC. The increase in the internal energy of the gas in kilo - joules is: (R=8.30Jmol−1 K−1)

5 moles of Hydrogen $(γ = \frac{7}{5})$ initially at S.T.P. are compressed adiabatically so that its temperature becomes $400^{o} C$ . The increase in the internal energy of the gas in kilo - joules is: $(R = 8.30 J m o l^{- 1} K^{- 1})$