Question

A motor-car tyre has a pressure of 2 atmosphere at 27⁰C. It suddenly bursts. If (C_p/C_v) = 1.4 for air, find the resulting temperature –

• A. 57⁰C
• B. 27⁰C
• C. –27⁰C
• D. 246⁰C

Answer 1

Answer: (C)

–27⁰C

Answer 2

PV^g = const. ; P $\left( \frac { \mathrm { T } } { \mathrm { P } } \right) ^ { \gamma }$ = const.

P^1–g T^g = const. ; P^1–g µ T^–g

P^–0.4 µ T^–1.4 ; P µ $\mathrm { T } ^ { \frac { 1.4 } { 0.4 } }$

or P µ T^7/2 or $\frac { \mathrm { T } _ { 2 } } { \mathrm {~T} _ { 1 } } = \left( \frac { \mathrm { P } _ { 2 } } { \mathrm { P } _ { 1 } } \right) ^ { 2 / 7 }$

or T₂= $\left( \frac { \mathrm { P } _ { 2 } } { \mathrm { P } _ { 1 } } \right) ^ { 2 / 7 }$ T₁=$\left( \frac { 1 } { 2 } \right) ^ { 2 / 7 }$300K = $\frac { 300 \mathrm {~K} } { 2 ^ { 2 / 7 } }$ < T₁

Temperature decreases in adiabatic expansion.