Question

91 2.0 1.0 0 0.2 0.4 Volume, m3 0.6 0.8

A gas expands along path ABC. The work done by the gas in this expansion is 7.0 * 10^5 J

It shows part of a cycle using one mole of a monatomic, ideal gas as the working fluid.

Answer 1

92) Heat capacity at constant volume is given by:

$Q = nc_{v}\bigtriangleup T$

Given n=1 (no. of moles), and $c_{v} = \frac{3}{2}R$ for a monoatomic ideal gas and heat capacity is defined for temperature change of $\bigtriangleup T =1$

So,

$Heat\ capacity = \frac{3}{2}R$

93) Heat capacity at constant pressure is given by:

$Q = nc_{p}\bigtriangleup T$

Given n=1 (no. of moles), and $c_{p} = \frac{5}{2}R$ for a monoatomic ideal gas and heat capacity is defined for temperature change of $\bigtriangleup T =1$

So,

$Heat\ capacity\ (constant\ pressure) = \frac{5}{2}R$

94) change in internal energy is given by:

$\bigtriangleup E_{int} = nc_{v}\bigtriangleup T$

So from A to A, the pressure and volume of the gas will be same and so the tempereture acoording to ideal gas law,

So, $\bigtriangleup T = 0$

and hence $\bigtriangleup E_{int} = 0$

95) acoording to ideal gas law,

$T \propto PV$

So, $\frac{T_{A}}{T_{C}} = \frac{P_{A}V_{A}}{P_{C}V_{C}}$

$\frac{T_{A}}{T_{C}} = \frac{2\times 10^{6}\times 0.4}{1\times 10^{6}\times 0.8}$

$\Rightarrow \frac{T_{A}}{T_{C}} = 1$

$\Rightarrow T_{A}= T_{C}$

So, temperature at A and C will be same

96) Q at pressure = 2 * 10⁶ ​Pascal will be

$Q_{2} = nC_{p}\bigtriangleup T$

$\Rightarrow Q_{2} = nC_{p}\frac{\bigtriangleup (PV)}{nR}$

$\Rightarrow Q_{2} = n\times \frac{3}{2}R\times \frac{0.2\times 10^{6}}{nR}$

$\Rightarrow Q_{2} = 0.3\times10^{6}$

Since, ​ So direction is inside the system