Question

1.75 L of a gas at 127°C and 5 atm is converted to 2.875 L at 85°C and 1 atm. The % of the gas that had escaped is
is
1) 11.11% 2) 23.2%
3) 54.3%
4) 63.5%​

Answer 1

Given:

1.75 L of a gas at 127°C and 5 atm is converted to 2.875 L at 85°C and 1 atm.

To find:

% of gas escaped.

Calculation:

Considering the gas to be Ideal ;

$\sf{\dfrac{(P1)(V1)}{(T1)} = \dfrac{(P2)(V2)}{(T2)} }$

$\sf{ = > \dfrac{(5)(1.75)}{(273 + 127)} = \dfrac{(1)(V2)}{(273 + 85)} }$

$\sf{ = > \dfrac{(5)(1.75)}{(400)} = \dfrac{(1)(V2)}{(358)} }$

$\sf{= > \: V2 = 7.83 \: litres}$

So, amount of gas escaped

∆V = 7.83 - 2.875 = 4.955 L

Therefore, percentage escape

$\rm{\Delta V\% = \dfrac{4.955}{7.83} \times 100\%}$

$\rm{ = > \Delta V\% = 63.3\%}$

$\rm{ = > \Delta V\% \approx 63.5\%}$

So, final answer is:

$\boxed{ \large{ \red{ \rm{\Delta V\% \approx 63.5\%}}}}$

Answer 2

Explanation:

Given:

1.75 L of a gas at 127°C and 5 atm is converted to 2.875 L at 85°C and 1 atm.

To find:

% of gas escaped.

Calculation:

Considering the gas to be Ideal ;

\sf{\dfrac{(P1)(V1)}{(T1)} = \dfrac{(P2)(V2)}{(T2)} }

(T1)

(P1)(V1)

=

(T2)

(P2)(V2)

\sf{ = > \dfrac{(5)(1.75)}{(273 + 127)} = \dfrac{(1)(V2)}{(273 + 85)} }=>

(273+127)

(5)(1.75)

=

(273+85)

(1)(V2)

\sf{ = > \dfrac{(5)(1.75)}{(400)} = \dfrac{(1)(V2)}{(358)} }=>

(400)

(5)(1.75)

=

(358)

(1)(V2)

\sf{= > \: V2 = 7.83 \: litres}=>V2=7.83litres

So, amount of gas escaped

∆V = 7.83 - 2.875 = 4.955 L

Therefore, percentage escape

\rm{\Delta V\% = \dfrac{4.955}{7.83} \times 100\%}ΔV%=

7.83

4.955

×100%

\rm{ = > \Delta V\% = 63.3\%}=>ΔV%=63.3%

\rm{ = > \Delta V\% \approx 63.5\%}=>ΔV%≈63.5%

So, final answer is:

\boxed{ \large{ \red{ \rm{\Delta V\% \approx 63.5\%}}}}

ΔV%≈63.5%