Question

A flask was heated from 27°C to 227°C at constant pressure. Calculate the volume of the flask if 0.1 $dm^{3}$ of air measured at 227°C was expelled from the flask ?

Answer 1

From the given

T = 27°C = 27 + 273 = 300 K

${ T }_{ 1 }$= 227°C = 227 + 273 = 500 K

V = 0.1 ${ dm }^{ 3 }$ = 1 litre  

${ V }_{ 1 }$ = ?  

$At\quad constant\quad pressure\quad =\quad \frac { V }{ T } \quad =\quad \frac { { V }_{ 1 } }{ { T }_{ 1 } } \\\\ \Rightarrow \quad \frac { 1 }{ 300 } \quad =\quad \frac { { V }_{ 1 } }{ 500 } \\ \\\Rightarrow \quad { V }_{ 1 }\quad =\quad \frac { 500 }{ 300 } \quad \\\\ \Rightarrow \quad { V }_{ 1 }\quad =\quad 1.66\quad L\\\\ Volume\quad of\quad air\quad escaped\quad out\quad =\quad 1.66\quad -\quad 1\\\\ =\quad 0.66\quad L\\ \\V\quad =\quad 66\quad mL$

The volume of the flask is 66 mL

Answer 2

Answer:

T = 27°C = 27 + 273 = 300 K

{ T }_{ 1 }T

1

= 227°C = 227 + 273 = 500 K

V = 0.1 { dm }^{ 3 }dm

3

= 1 litre

{ V }_{ 1 }V

1

= ?

\begin{gathered}At\quad constant\quad pressure\quad =\quad \frac { V }{ T } \quad =\quad \frac { { V }_{ 1 } }{ { T }_{ 1 } } \\\\ \Rightarrow \quad \frac { 1 }{ 300 } \quad =\quad \frac { { V }_{ 1 } }{ 500 } \\ \\\Rightarrow \quad { V }_{ 1 }\quad =\quad \frac { 500 }{ 300 } \quad \\\\ \Rightarrow \quad { V }_{ 1 }\quad =\quad 1.66\quad L\\\\ Volume\quad of\quad air\quad escaped\quad out\quad =\quad 1.66\quad -\quad 1\\\\ =\quad 0.66\quad L\\ \\V\quad =\quad 66\quad mL\end{gathered}

Atconstantpressure=

T

V

=

T

1

V

1

⇒

300

1

=

500

V

1

⇒V

1

=

300

500

⇒V

1

=1.66L

Volumeofairescapedout=1.66−1

=0.66L

V=66mL

Explanation:

The volume is 66