One mole of an ideal gas (γ = 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 = 8.3 J-1 mole-1 K-1)

Question

One mole of an ideal gas (γ = 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 = 8.3 J-1 mole-1 K-1) (A) - 166 J (B) 166 J (C) - 168 J (D) 168 J

Tardigrade Admin · Accepted Answer

ÎU = CV ÎT. Now CP-CV or (CP/CV)-1=(R/CV) or CV=(R/&gamma;-1') (∵ &gamma;=(CP/CV).) Hence Δ u=(RΔ T/(&gamma;-1))=(8.3×8/(1.4-1))=166 J

Q. One mole of an ideal gas $(γ = 1.4)$ is adiabatically compressed so that its temperature rises from $2 7^{°} C$ to $3 5^{°} C$ . The change in the internal energy of the gas is (given $R = 8.3 J^{- 1} m o l e^{- 1} K^{- 1}$ )

$- 166 J$

$166 J$

$- 168 J$

$168 J$

$Δ U = C_{V} Δ T .$
Now
$C_{P} - C_{V}$ or $\frac{C _{P}}{C _{V}} - 1 = \frac{R}{C _{V}}$
or $C_{V} = \frac{R}{γ - 1 ^{^{'}}} (∵ γ = \frac{C _{P}}{C _{V}} .)$
Hence $Δ u = \frac{R Δ T}{( γ - 1 )} = \frac{8.3 \times 8}{( 1.4 - 1 )} = 166 J$

Q. One mole of an ideal gas (γ=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=8.3J−1mole−1K−1)

−166J

166J

−168J

168J