ThermodynamicsHard
Question
Two moles of an ideal gas [Cv,m (/JK−1 mol−1) = 20 + 0.01T(/K)] is heated at constant pressure from 27°C to 127°C. The amount of heat absorbed by the gas is
Options
A.1662.8 J
B.4700 J
C.6362.8 J
D.3037.2 J
Solution
$q_{p} = n.\int_{T_{1}}^{T_{2}}{C_{p,m}.dT} = 2 \times \int_{T_{1} = 300\text{ K}}^{T_{2} = 400\text{ K}}\left( 28.314 + 0.01\text{ T} \right).dT$
$= 2\left\lbrack 28.314\left( T_{2} - T_{1} \right) + \frac{0.01}{2}\left( T_{2}^{2} - T_{1}^{2} \right) \right\rbrack = 6362.8\text{ J}$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
The plot of $\log_{10}\text{ }K$ vs $\frac{1}{\text{ }T}$ gives a straight line. The intercept and slope respectively ar...For the reaction, X2O4(l) → 2XO2(g) ᐃU = 2.1 kcal, ᐃS = 20 cal K-1 at 300 K Hence, ᐃG is...A schematic plot of In Keq versus inverse of temperature for a reaction is shown below The reaction must be...Heat liberated by a given amount of an ideal gas undergoing reversible isothermal process is 1200 cal at 300 K. What is ...The volume of a system becomes twice its original volume on the absorption of 300 cal of heat. The work done on the surr...