Tardigrade - CET NEET JEE Exam App
© 2022 Tardigrade®. All rights reserved Text Solution Solution : Let the initial volume be x , `V_1=x` <br> `V_2` = ? <br> `T_1`= 273 K <br> `T_2=273^@C` = 273+273 =546 K <br> According to Charles.s law, `V_1/T_1=V_2/T_2` <br> `x/273=V_2/546` <br> `V_2=(546xx x)/273` <br> `V_2`= 2x <br> Thus, it is proven that the volume of any gas at `273^@C` is twice its volume at 273 K, at constant pressure. Answer VerifiedRight on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App 2 |