Question

volume of a gas at stp is 10 dm3 at what temp it's volume will be 30dm3
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Answer 1

Answer:

YUP Since the temperature remains constant, Boyle's law is applicable

This means that :

P  

1

​

V  

1

​

=P  

2

​

V  

2

​

 

Now since volume is halved, pressure becomes twice because product, PV = constant

Explanation:

Answer 2

Answer: The temperature at which volume of gas will be $30 dm^{3}$ is $819K$.  

Explanation:

Given: At STP conditions ($1 atm$ pressure and $273 K$ temperature)

Volume of gas, $V_{1}= 10 dm^{3}$, $T_{1} = 273 K$

Volume of gas, $V_{2} = 30 dm^{3}$

To find: Temperature $(T_{2} )$ at which volume of  gas is $V_{2}$.

Solution: As the STP conditions are given;

At constant pressure, by applying Charles's law to calculate $T_{2}$:

$\frac{V_{2} }{V_{1} } = \frac{T_{2} }{T_{1} }$

On putting the values

$\frac{30 dm^{3} }{10 dm^{3} } = \frac{T_{2} }{273 K}$

$T_{2} = 819 K$

At $819 K$, volume of gas will be $30 dm^{3}$.

#SPJ2