Question

Q.1 in a constant temperature process 70 moles of
an ideal gas at temperature 354 K attains a final
volume V, Work input during this process is
206 kJ. Initial volume V, of the gas
approximately satisfies the following relation
(e is the base of natural logarithm)
(a) V = V2 (b) V = eV2
Subject matter to MADE EASY Publications, New Del
(o) in
(d) V = In V2
Q.5 C​

Answer 1

Answer:

please mark me bra list

Explanation:

Given,

P∝V

−2

n=2

T

1

=300K

T

2

=400K

PV

2

=const. . . . . .(1)

γ=2 (PV

γ

=constant)

The above equation shows that the process is adiabatic process,

The work done in adiabatic process in thermodynamics is given by

W=nR

γ−t

(T

1

−T

2

)

W=nR

2−1

(300−400)

W=−100nR=−2×100R

W=−200R

The correct option is B.

Answer 2

Answer:

Initial volume satisfies the equation $V = \frac{V_{2} }{e }$

Explanation:

Given, moles of an ideal gas, $n=70$

temperature,  $T=354 K$

and work input during this process,  $W=206 kJ$

Since there is change in volume at temperature remaining constant, it is an isothermal process.

Work done in isothermal process is,

                                   $W=2.3026nRTlog_{10}\frac{V_{2} }{V_{1} }$

                $\implies 206\times 10^{3} =2.3026\times 70 \times 8.31 \times 354 log_{10}\frac{V_{2} }{V} }$

   $\implies \frac{206\times 10^{3}}{2.3026\times 70 \times 8.31 \times 354} = log_{10}\frac{V_{2} }{V }$

                       $\implies \frac{20600}{47305} = log_{10}\frac{V_{2} }{V }$

                        $\implies 0.43 = log_{10}\frac{V_{2} }{V }$

                     $\implies 10^{0.43} = \frac{V_{2} }{V }\implies 2.7 = \frac{V_{2} }{V }$  

                                                   $\implies e = \frac{V_{2} }{V }\implies V = \frac{V_{2} }{e }$

Hence, initial volume, $V = \frac{V_{2} }{e }$