One mole of an ideal gas Y -1-4 is adiabatically compressed so that its temperature rises from 27- C to 35- C- The change in the internal energy of the gas is given R -80 - J - mole - K A- -166 JB- 166 JC- -168 JD- 168 J

The correct option is B 166 J Δ U = C_v Δ T. Now C_p C_v = R or C_p/C_v 1 = R/C_v or C_v = R/γ 1, where γ = C_p/C_v. Hence Δ U= R Δ T/(γ 1) = 8.3 × 8/(1 ...

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Standard XII

Physics

Thermodynamic Processes

One mole of a...

Question

One mole of an ideal gas $(γ = 1.4)$ is adiabatically compressed so that its temperature rises from $27^{o} C t o 35^{o} C$ . The change in the internal energy of the gas is (given R = 80. J / mole / K)

- 166 J

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166 J

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- 168 J

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168 J

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Solution

The correct option is B
166 J

$Δ U = C_{v} Δ T . N o w C_{p} - C_{v} = R o r \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} o r C_{v} = \frac{R}{γ - 1}, w h e r e γ = \frac{C_{p}}{C_{v}} . H e n c e Δ U = \frac{R Δ T}{(γ - 1)} = \frac{8.3 \times 8}{(1.4 - 1)} = 166 J$

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One mole of an ideal gas (γ=1.4) is adiabatically compressed so that its temperature rises from 27oC to 35oC. The change in the internal energy of the gas is (given R = 80. J / mole / K)

The correct option is B 166 J ΔU=CvΔT. Now Cp−Cv=R or CpCv−1=RCvor Cv=Rγ−1, where γ=CpCv. HenceΔU=RΔT(γ−1)=8.3×8(1.4−1)=166 J

One mole of an ideal gas $(γ = 1.4)$ is adiabatically compressed so that its temperature rises from $27^{o} C t o 35^{o} C$ . The change in the internal energy of the gas is (given R = 80. J / mole / K)

The correct option is B
166 J

$Δ U = C_{v} Δ T . N o w C_{p} - C_{v} = R o r \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} o r C_{v} = \frac{R}{γ - 1}, w h e r e γ = \frac{C_{p}}{C_{v}} . H e n c e Δ U = \frac{R Δ T}{(γ - 1)} = \frac{8.3 \times 8}{(1.4 - 1)} = 166 J$