sir ours is integrated sylabus

i am in class8

but now i go to class9

this si from fiitjee material

sir

Log in to add a comment

sir ours is integrated sylabus

i am in class8

but now i go to class9

this si from fiitjee material

sir

Log in to add a comment

The Brainliest Answer!

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

2. P2 / P1 = T2/ T1

10^6 / 0.25 * 10^6 = T2 / 300 K

=> T2 = 1200 K

It will blow up at 1200 K

============

3.

P1 = 5 atm T1 = 303 K V1 n1 m1

P1 V1 = n1 R T1

T2 = 293 K, n1, m1 same

V2 = 160% V1 = 1.6 V1 => P2 = P1 /1.6

n3 = n1 * 120% = 1.2 n1 V3 = V2 = 1.6 V1

T3 = ? P3 = P1

P3 V3 / [P1 V1 ] = n3 R T3 / [ n1 R T1 ]

1.6 = 1.2 * T3 / 303

T3 = 1.6 / 1.2 * 303 K

========================

4.

I think there is a misprint in the number : 10 deg C increase in temperature or 10% increase in pressure.... Increase of 10 deg. C cannot increase pressure by 10%.

m = 12 gm T1 = 273+t deg K V1 = v litre P1 = 1 atm

M = 120 => n = 0.1 number of moles

*if the increase of temperature is: 100 deg C*

* * T2 = 373+t deg K V2 = v litre P2 = 1.10 atm

P2/P1 = T2 / T1 => 1.10 (273 + t) = 373 + t

=> t = 10 * 72.7 = 727 deg C

if the increase of pressure is 1 % only and increase of temperature is 10 degC,

P2/P1 = 1.01 = 283+t / (273+t)

t = 727 deg C

=================================

then calculate V as:

v = n * R (273+t) / P1 as all quantities are known

10^6 / 0.25 * 10^6 = T2 / 300 K

=> T2 = 1200 K

It will blow up at 1200 K

============

3.

P1 = 5 atm T1 = 303 K V1 n1 m1

P1 V1 = n1 R T1

T2 = 293 K, n1, m1 same

V2 = 160% V1 = 1.6 V1 => P2 = P1 /1.6

n3 = n1 * 120% = 1.2 n1 V3 = V2 = 1.6 V1

T3 = ? P3 = P1

P3 V3 / [P1 V1 ] = n3 R T3 / [ n1 R T1 ]

1.6 = 1.2 * T3 / 303

T3 = 1.6 / 1.2 * 303 K

========================

4.

I think there is a misprint in the number : 10 deg C increase in temperature or 10% increase in pressure.... Increase of 10 deg. C cannot increase pressure by 10%.

m = 12 gm T1 = 273+t deg K V1 = v litre P1 = 1 atm

M = 120 => n = 0.1 number of moles

P2/P1 = T2 / T1 => 1.10 (273 + t) = 373 + t

=> t = 10 * 72.7 = 727 deg C

if the increase of pressure is 1 % only and increase of temperature is 10 degC,

P2/P1 = 1.01 = 283+t / (273+t)

t = 727 deg C

=================================

then calculate V as:

v = n * R (273+t) / P1 as all quantities are known