5 moles of hydrogen (γ= 7 / 5) intially at S.T.P. are compressed adiabatically so that its temperature becomes 4000C. The increase in the internal energy of the gas in kilo-J is (R=8.30Jmole–1k–1)

n=5 mole ; γ=75T1=273K;T2=673KdU=?R=8.30J/ mole kCV=1n×(dQ)VdT; here (dQ)V=dUdU=nCVdT=nR/γ−1dTdU=5×8.3×4007/5−1=25×83×402dU=41500J=41.50KJ

5 moles of hydrogen (γ= 7 / 5) intially at S.T.P. are compressed adiabatically so that its temperature becomes 4000C. The increase in the internal energy of the gas in kilo-J is (R=8.30Jmole–1k–1)

n=5 mole ; γ=75T1=273K;T2=673KdU=?R=8.30J/ mole kCV=1n×(dQ)VdT; here (dQ)V=dUdU=nCVdT=nR/γ−1dTdU=5×8.3×4007/5−1=25×83×402dU=41500J=41.50KJ

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5 moles of hydrogen ( $γ =$ 7 / 5) intially at S.T.P. are compressed adiabatically so that its temperature becomes 400⁰C. The increase in the internal energy of the gas in kilo-J is (R=8.30Jmole^–1k^–1)

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21.55

41.50

65.55

80.55

answer is B.

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Detailed Solution

$\begin{array}{l} n = 5 mole; γ = \frac{7}{5} \\ T_{1} = 273 K; T_{2} = 673 K \\ dU = ? R = 8 .30 J / mole k \\ C_{V} = \frac{1}{n} \times \frac{(dQ)_{V}}{dT}; \\ here (dQ)_{V} = dU \\ dU = {nC}_{V} dT = nR / γ - 1 dT \\ dU = \frac{5 \times 8 .3 \times 400}{7 / 5 - 1} = \frac{25 \times 83 \times 40}{2} \\ dU = 41500 J = 41 .50 KJ \end{array}$