Question

A gas is collected at 37°C and 10 atm pressure. It is desired to reduce its volume to 1/5 th of the original volume at a pressure of 12 atm. To what temperature should it be cooled?​

Answer 1

Answer:

Hope this will be helpful

Answer 2

Answer:

The final temperature measured is $-198.72\textdegree C$.

Explanation:

Given,

The initial pressure, P₁ = $10atm$

The final pressure, P₂ = $12atm$

The initial temperature, T₁ = $37\textdegree C$

Convert °C in K

• $1\textdegree C = 273.15 K$
• $37\textdegree C =( 37 + 273.15) K$ = $310.15K$

Let the initial volume = V₁

Therefore, the final volume = $\frac{V_{1} }{5}$

The final temperature, T₂ =?

As we know,

• According to the combined gas law;
• $\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

After putting the given values in the equation, we get:

• $\frac{10\times V_{1} }{310.15 } = \frac{12\times V_{1} }{5\times T_{2} }$
• $50 T_{2} = 3721.8$
• $T_{2} = 74.43K$

Convert K in °C

• $\textdegree C = K - 273.15$
• $\textdegree C = (74.43 - 273.15)\textdegree C$ = $-198.72\textdegree C$

Hence, the final temperature, T₂ = $-198.72\textdegree C$.