At constant volume and 27-o-C the pressure of a gas is 2 atmosphere- At 327-o-C- the pressure of the gas in an atmosphere will be-1-5241

Question

Toppr Admin · Accepted Answer

Given- P1-2 atmT1-300 KP2-T2-600 KSince volume is constant-xA0-Charles law is applicable- P-x221D-TSo- P1T1-P2T2Putting all the given values- we get P2-xA0- - 4 atmOption D is correct-

At constant volume and $27^{o} C$ the pressure of a gas is 2 atmosphere. At $327^{o} C$ , the pressure of the gas in an atmosphere will be:
1.5
2
4
1

Given, $P_{1} = 2$ $a t m$
$T_{1} = 300$ $K$
$P_{2} = ?$
$T_{2} = 600$ $K$

Since volume is constant,
Charles law is applicable, $P \propto T$

So, $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

Putting all the given values, we get $P_{2}$ = $4$ $a t m$
Option D is correct.

At constant volume and 27oC the pressure of a gas is 2 atmosphere. At 327oC, the pressure of the gas in an atmosphere will be:1.5241

Given, P1=2 atmT1=300 KP2=?T2=600 KSince volume is constant, Charles law is applicable, P∝TSo, P1T1=P2T2Putting all the given values, we get P2 = 4 atmOption D is correct.

At constant volume and $27^{o} C$ the pressure of a gas is 2 atmosphere. At $327^{o} C$ , the pressure of the gas in an atmosphere will be:
1.5
2
4
1

Given, $P_{1} = 2$ $a t m$
$T_{1} = 300$ $K$
$P_{2} = ?$
$T_{2} = 600$ $K$

Since volume is constant,
Charles law is applicable, $P \propto T$

So, $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

Putting all the given values, we get $P_{2}$ = $4$ $a t m$
Option D is correct.