Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 4 (Q.No. 8)
8.
A car tyre of volume 0.057 m3 is inflated to 300 kPa at 300 K. After the car is driven for 10 hours, the pressure in the tyre increases to 330 kPa. Assume air is an ideal gas and Cv for air is 21 J/mole.K. The change in the internal energy of air in the tyre in J/mole is
Discussion:
1 comments Page 1 of 1.
Nader said:
8 years ago
pv=nrt.
n = (3*10^5 * 0.057) / (8.317 * 300) = 6.853.
dU= Cv * dT,
dU = 21* (T2-300),
T2= P2 V2 / n R = (3.3*10^5 * 0.057) / 6.853 * 8.314.
T2=330 K.
dU= 21 * (330-300) = 630 J/Mole.
n = (3*10^5 * 0.057) / (8.317 * 300) = 6.853.
dU= Cv * dT,
dU = 21* (T2-300),
T2= P2 V2 / n R = (3.3*10^5 * 0.057) / 6.853 * 8.314.
T2=330 K.
dU= 21 * (330-300) = 630 J/Mole.
(2)
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