An ideal gas initially at pressure 1 , bar is | vedclass

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(A) For an ideal gas,the combined gas law is given by the equation: $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$.Given values are:Initial p

Vedclass · Accepted Answer

$2.625$

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(A) For an ideal gas,the combined gas law is given by the equation: $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$.Given values are:Initial pressure $P_{1} = 1 \, bar$Initial volume $V_{1} = 30 \, m^{3}$Initial temperature $T_{1} = 320 \, K$Final volume $V_{2} = 10 \, m^{3}$Final temperature $T_{2} = 280 \, K$Substituting these values into the equation:$\frac{1 	imes 30}{320} = \frac{P_{2} 	imes 10}{280}$Solving for $P_{2}$:$P_{2} = \frac{1 	imes 30 	imes 280}{320 	imes 10}$$P_{2} = \frac{3 	imes 280}{320} = \frac{840}{320} = 2.625 \, bar$.Thus,the final pressure of the gas is $2.625 \, bar$.

An ideal gas initially at pressure $1 \, bar$ is being compressed from $30 \, m^{3}$ to $10 \, m^{3}$ volume and its temperature decreases from $320 \, K$ to $280 \, K$. Find the final pressure of the gas (in $bar$).

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