An ideal gas having initial pressure P , volume V and temperature T is allowed to expand adiabatically until its volume becomes 4 ~V , while its temperature falls to T/2 . If the work done by the gas during the expansion is α PV , the value of α is

Correct answer: (B) 1.0. For adiabatically process T_1 ~V_1^γ-1=T_2 ~V_2^γ-1 TV^γ-1=T/2(4 ~V)^γ-1 2=(2)^2 γ-2 γ=3/2 2 γ-2=1 Work done by the gas W=(nR (T_2-T_1 ))/(γ-1)=(nR (T-T/2 ))/(3/2-1) =(nR T/2)/(1 / 2) =nRT= PV So, α=1

Q139 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Physics

An ideal gas having initial pressure $\mathrm{P}$, volume $\mathrm{V}$ and temperature $\mathrm{T}$ is allowed to expand adiabatically until its volume becomes $4 \mathrm{~V}$, while its temperature falls to $\frac{\mathrm{T}}{2}$. If the work done by the gas during the expansion is $\alpha \mathrm{PV}$, the value of $\alpha$ is

A
1.25
Correct answer
B
1.0
C
1.50
D
2.0

Solution

For adiabatically process $$ \begin{array}{ll} \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} & \\ \mathrm{TV}^{\gamma-1}=\frac{\mathrm{T}}{2}(4 \mathrm{~V})^{\gamma-1} & \\ 2=(2)^{2 \gamma-2} & \gamma=\frac{3}{2} \\ 2 \gamma-2=1 & \end{array} $$ Work done by the gas $$ \begin{aligned} & \mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{T}_2-\mathrm{T}_1\right)}{\gamma-1}=\frac{\mathrm{nR}\left(\mathrm{T}-\frac{\mathrm{T}}{2}\right)}{\frac{3}{2}-1} \\ & =\frac{\mathrm{nR} \frac{\mathrm{T}}{2}}{1 / 2} \\ & =\mathrm{nRT}=\text { PV So, } \alpha=1 \end{aligned} $$

Hence, the correct answer is option: 'B'.

Solution

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