A gasoline engine takes in 5 moles of air at 20- C and 1 atm- and compresses it adiabatically to 1-10 th of the original volume- Find the change in internal energy- Assume air to be diatomicA- 92 kJB- 46 kJC- 82 kJD- 23 kJ

The correct option is B 46 kJ P_1 = 1 atm, n = 5, T_1 = 293 K V_2=V_1/10 Now, P_1V_1^x=P_2V_2^x P_2 = 25.12 atm Similarly, T_1V_1^x 1 = T_2V_2^x 1 T_2 = 73 ...

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Standard XII

Physics

Temperature

A gasoline en...

Question

A gasoline engine takes in 5 moles of air at $20^{\circ}$ C and 1 atm, and compresses it adiabatically to ${\frac{1}{10}}^{t h}$ of the original volume. Find the change in internal energy. (Assume air to be diatomic)

92 kJ

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23 kJ

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46 kJ

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82 kJ

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Solution

The correct option is C
46 kJ

$P_{1} = 1$ atm, $n = 5, T_{1} = 293 K$

$V_{2} = \frac{V_{1}}{10}$

Now, $P_{1} V_{1}^{x} = P_{2} V_{2}^{x}$

$P_{2} = 25.12$ atm

Similarly, $T_{1} V_{1}^{x - 1} = T_{2} V_{2}^{x - 1}$

$T_{2} = 736 K$

Work done $= \frac{n R (T_{1} - T_{2})}{x - 1} = - 46 k J$

$Δ U = Δ H - W = 0 - W = 46 k J$

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A gasoline engine takes in 5 moles of air at 20∘ C and 1 atm, and compresses it adiabatically to 110th of the original volume. Find the change in internal energy. (Assume air to be diatomic)

The correct option is C 46 kJ P1=1 atm, n=5,T1=293 K V2=V110 Now, P1Vx1=P2Vx2 P2=25.12 atm Similarly, T1Vx−11=T2Vx−12 T2=736 K Work done =nR(T1−T2)x−1=−46 kJ Δ U=Δ H−W=0−W=46 kJ

A gasoline engine takes in 5 moles of air at $20^{\circ}$ C and 1 atm, and compresses it adiabatically to ${\frac{1}{10}}^{t h}$ of the original volume. Find the change in internal energy. (Assume air to be diatomic)