Question

15. A sample of a gas occupies 10L at 240°C under a pressure
of 80 kPa. At what temperature would the gas occupy 20L,
if we increase the pressure to 100.0 kPa ? (1009.5°C]​

Answer 1

Explanation:

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Answer 2

★Given :

• A sample of a gas occupies 10L .
• Temperature =  240°C  
• Pressure = 80 kPa
• Pressure is increased to 100.0kPa
• Volume in the second case = 20L

★To find :

• The temperature in the second case.

★Solution :  

Converting Celsius to kelvin :  

Given T = 240°C

→K = C + 273

→ 240 + 273

→ 513 k  

We have :

• $\sf V_1 = 10L$
• $\sf P_1 = 80\ kPa$
• $\sf T_1 = 513\ k$
• $\sf V_2 = 20L$
• $\sf P_2 = 100\ kPa$
• $\sf T_2 = \ ?$

Now,Using the general gas equation,

$\displaystyle \leadsto \boxed{\sf \frac{P_1V_1}{T_1}=\frac{P_2 V_2}{T_2} }$

Substituting the values,

$\implies \sf T_2 = \dfrac{P_2V_2}{P_1V_1} (T_1)$

$\implies \sf \dfrac{100\times 20 }{80\times 10 } (513)$

$\implies \sf \dfrac{5}{2} \times 513$

$\implies \sf 2.5\times 513$

$\implies \sf 1282.5 k$

$\implies \boxed{ 1009.5 \degree C}$

Therefore,the temperature is 1009.5°.

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