Question

One mole of an ideal gas undergoes a process P =$\frac { \mathrm { P } _ { 0 } } { 1 + \left( \frac { \mathrm { V } _ { 0 } } { \mathrm {~V} } \right) ^ { 2 } }$

Here, P₀ and V₀ are constants. Change in temperature of the gas when volume is changed from V = V₀ to V = 2V₀ is –

• A. –
• B. $\frac { 11 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { 10 \mathrm { R } }$
• C. –
• D. P₀V₀

Answer 1

Answer: (B)

$\frac { 11 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { 10 \mathrm { R } }$

Answer 2

At V = V₀, P =

\ T_i == (n = 1)

and at V = 2V₀, P =$\frac { 4 \mathrm { P } _ { 0 } } { 5 }$

\ T_f =$\frac { 8 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { 5 \mathrm { R } }$

\ DT = T_f – T_i =$\frac { 11 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { 10 \mathrm { R } }$