Question

5 l of a gas at a pressure of 2 atm and a temperature of 27 degree celsius is heated to 227 degree celsius and pressure is rice 6 atm . what is the volume under this condition meritination for 1

Answer 1

Explanation:

p1= 2atm. P2=6atm

T1=300K. V1=5L

T1=273+227=500K. V2=?

(P1V1) = (P2V2)

T1. T2

THEREFORE

V2=P1V1T2 = 2×5×500 = 2.77L

TIP2. 300×6

Answer 2

Given: $P_{1} =2atm$, $T_{1} =27^{o} C=300K$, $V_{1} =5V$

          $P_{2} =6atm$, $T_{1} =227^{o} C=500K$

To find: $V_{2}$

Step-by-step explanation:

Step 1 of 2

The ideal gas equation is $PV=nRT$.

It can be written as: $nR=\frac{PV}{T}$

For a gas the gas constant and number of moles are constant it can be written as:

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

Step 2 of 2

Substituting the values we will get:

$\frac{2*5}{300} =\frac{6*V_{2} }{500} \\V_{2} =\frac{2*5*500}{300*6} \\V_{2} =\frac{5000}{1800} \\V_{2} =2.77L$

The volume under this condition is 2.77L.