An insulated cylinder of volume is filled with a gas at and 2 atmospheric pressure. Then the gas is compressed making the final volume as while allowing the temperature to rise to . The final pressure is __________ atmospheric

Question

An insulated cylinder of volume $60 {cm}^{3}$ is filled with a gas at $27 ° C$ and 2 atmospheric pressure. Then the gas is compressed making the final volume as $20 {cm}^{3}$ while allowing the temperature to rise to $77 ° C$ . The final pressure is __________ atmospheric pressure. [2026]

Smart Achievers · Accepted Answer

(7)

$P V = n R T$

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

$\frac{2 \times 10^{5} \times 60}{300} = \frac{P_{2} \times 20}{350}$

$P_{2} = \frac{2 \times 10^{5} \times 7 \times 3}{6}$

$P_{2} = 7 \times 10^{5} = 7 atm$

Solution

Q.
An insulated cylinder of volume $60 {cm}^{3}$ is filled with a gas at $27 ° C$ and 2 atmospheric pressure. Then the gas is compressed making the final volume as $20 {cm}^{3}$ while allowing the temperature to rise to $77 ° C$ . The final pressure is __________ atmospheric pressure. [2026]

Ans.
(7)

$P V = n R T$

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

$\frac{2 \times 10^{5} \times 60}{300} = \frac{P_{2} \times 20}{350}$

$P_{2} = \frac{2 \times 10^{5} \times 7 \times 3}{6}$

$P_{2} = 7 \times 10^{5} = 7 atm$

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Solution

Q. An insulated cylinder of volume 60 cm3 is filled with a gas at 27°C and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20 cm3 while allowing the temperature to rise to 77°C. The final pressure is __________ atmospheric pressure. [2026]

Ans. (7) PV=nRT P1V1RT1=P2V2RT2 2×105×60300=P2×20350 P2=2×105×7×36 P2=7×105=7 atm

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Q.
An insulated cylinder of volume $60 {cm}^{3}$ is filled with a gas at $27 ° C$ and 2 atmospheric pressure. Then the gas is compressed making the final volume as $20 {cm}^{3}$ while allowing the temperature to rise to $77 ° C$ . The final pressure is __________ atmospheric pressure. [2026]

Ans.
(7)

$P V = n R T$

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

$\frac{2 \times 10^{5} \times 60}{300} = \frac{P_{2} \times 20}{350}$

$P_{2} = \frac{2 \times 10^{5} \times 7 \times 3}{6}$

$P_{2} = 7 \times 10^{5} = 7 atm$